Given a year and a month, I want to determine the last day of that month. For example, if the year is 2004 and the month is 2, then the last day is 29th because of leap year.
Calculating the last day of a month is not hard, but complicated by the leap year. If the year is a leap year, then February’s last day will be 29th instead of the usual 28th. My algorithm to find the last day of the month is simpler than that: take the first day of the next month and count backward by one day. Here is the code.
#!/usr/bin/env python import datetime def get_last_day_of_the_month(y, m): ''' Returns an integer representing the last day of the month, given a year and a month. ''' # Algorithm: Take the first day of the next month, then count back # ward one day, that will be the last day of a given month. The # advantage of this algorithm is we don't have to determine the # leap year. m += 1 if m == 13: m = 1 y += 1 first_of_next_month = datetime.date(y, m, 1) last_of_this_month = first_of_next_month + datetime.timedelta(-1) return last_of_this_month.day
The code above is easy enough to understand. The datetime.timedelta(-1) on line 22 basically says, “subtract one day.”


Card counting in blackjack is somehow a nightmare for those us who are a bit forgetful or are dealing with mathematical anxiety. But this article will overpower all your fears and lack of coherent strategy. A punter doesn’t actually need to remember how many aces, queens or 4s are dealt out of the shoe. As we know that in the intense game of blackjack, all the picture cards hold a value of 10 and aces may be treated as 1s or 11 as per the hand played. So, more the aces and 10 value cards are lasting with the shoe more it is beneficial for the player. It influences the odds of winning. So, truly it is not mere keeping the track of number of cards dealt but it is actually hi low card counting i.e. the ratio of low value cards to high value cards.
Let’s simplify blackjack card counting practice by understanding few basic steps.
-> Assignment of values to cards:
Card of 2 to 6 holds a value = +1
Card of 7 to 9 holds a value = 0 and,
Card of value 10 to ace holds a value = -1.
-> Keeping a Running Count
These values signify that with each card dealt, a player will not be keeping the count of that card, rather will either be adding 1, subtracting 1 or doing nothing as per the above assigned values. This is done card after card and for each and every round until the dealer shuffles the card from the shoe. This strategy was quite helpful with the single deck game of blackjack. Higher the count, more are the odds of winning for the player. The high value cards pay out well and increases the chances of dealer going “bust” or over 21.
-> Keeping a True Count for multiple deck game of blackjack.
Maintaining a true count actually means to calculate a running count per deck. Casinos are smart enough nowadays to play with more than one deck. So, a shrewd player actually needs to calculate the true count by dividing the running count by the number of remaining decks . Maintaining this true count can take the bet to your advantage and you can place your strategic moves for much more cash.
-> Playing Deviations
Experts suggest to raise your bets as the true count rises and lower down the bets with the falling true count. Even a neutral true count doesn’t favour the player much.
Card counting is also about speed and accuracy. So, rather than each time going with calculations of getting a 4 (+1) and a jack (-1), resulting in a running count of “0”, take the card counting in pairs. Just keep in mind that whenever you get a high card and a low card, they cancel each other out and will simply be a “0”. This way, you will be a hands on expert in Las Vegas style card counting. Gradually a smart player can master other variants of card counting as well, like Omega, KO and halves. So here’s wishing you the best of both the worlds with luck and strategy.


Roulette Wheel Numbers
Roulette Wheel has the numbers 0-36 in three colours: green, red and black. Zero is Green. Half of the numbers 1-36 are Red, and half are Black.
Red numbers of the Roulette Wheel are 1, 3, 5, 7, 9, 12, 14, 16, 18, 19, 21, 23, 25, 27, 30, 32, 34 and 36.
Black numbers of the Roulette Wheel are 2, 4, 6, 8, 10, 11, 13, 15, 17, 20, 22, 24, 26, 28, 29, 31, 33 and 35.
The Green number of the Roulette Wheel is zero. European Roulette only has one green zero. American Roulette also has a green double zero (00). Some online Roulette games also have triple zeros (000).
The sum of all the Red numbers is 332 and Black 334. Summed up, all the numbers of the Wheel are 666. This gives Roulette its nickname, “the devil’s game”.
How it Works
The mechanical Wheel (in a real-life brick-and-mortar casino) is made of wood and metal. The Wheel rests on precision bearings, carefully selected to keep the spin results random (non-biased). The ball, originally made of ivory, is nowadays a special plastic mix. The construction of a wheel is an extremely detailed process, as any imperfection would lead to the results not being random enough. Only a few companies make Roulette wheels for top-end casinos, among them TCS John Huxley whose tables are often seen at live online casinos too.
Virtual Roulette Wheels are coded like any online game, but much work goes towards the game’s random number generator. This piece of coding ensures that the spin results are random. These RNGs are nowadays verified by special testing laboratories, where millions (or even billions) of results are drawn and then analyzed to ensure that the Wheel’s results are random. Game providers like Microgaming and Netent provide simulated games to online casinos.
Typical Roulette Wheel odds for getting a winning number on a European Roulette game are:
- Red: 48.6% (18 out of 37 numbers)
- Black: 48.6% (18 out of 37 numbers)
- Odds: 48.6% (18 out of 37 numbers)
- Even: 48.6% (18 out of 37 numbers)
- Low (Manque): 48.6% (18 out of 37 numbers)
- High (Passe): 48.6% (18 out of 37 numbers)
- Zero: 2.7% (1 out of 37 numbers)
- Street: 29.7% (3 out of 37 numbers)
The odds are slightly lower for double and triple zero games as there are 1-2 more non-winning numbers on the Wheel.
How to Calculate Odds
First, see how many numbers there are on the Wheel. This is either 37 (European single zero games), 38 (two zeros) or 39 (three zeros). Then divide the number of winning results by how many numbers there are on the Wheel.
For example, the “first dozen” bet has 12 winning numbers. Therefore its odds in an American Roulette game with two zero fields (0 and 00) are 12/38 = 0.31578 = 31.6%.
Roulette is a game of luck, so there are no working strategies for it. Casinos maintain the wheels daily, so they don’t favour any numbers over the others. Online casinos, on the other hand, use random number generators that are tested to be fully random; thus, you can’t predict what number will come up next.
Try it Yourself
You can try simulated Roulette Wheels on many sites. You can, of course, choose any online casino and try playing Roulette for free. Roulette is the most popular table game at casinos, so you’ll find at least a dozen different games at each casino.
The other option is to visit Roulette-Simulator.info, which features a great Roulette simulator you can play for free. You can play the simulator’s virtual game as long as you like and get yourself familiar with the ins and outs of Roulette Wheels. The game even keeps track of your results (if you so wish) so you can try to gamble yourself on the game’s high score list!
Last updated: 2023/03/14


JOSEPH MAZUR (born in the Bronx in 1942) is Professor Emeritus of Mathematics at Marlboro College, in Marlboro, Vermont.
He holds a Ph.D. in Mathematics from M.I.T., is a recipient of fellowships from the Guggenheim, Bogliasco, and Rockefeller Foundations, among others. His works have appeared in Nature, The New York Times, The Guardian, The Wall Street Journal, Slate, Science, and many other publications. He has been profiled in media venues such as NPR's "The Hidden Brain" and PRI's "Innovation Lab", CBS, the BBC, Vox, Radio Australia, Radio Ireland, and dozens of others. He is the author of Euclid in the Rainforest: Discovering Universal Truth in Mathematics; The Motion Paradox: The 2,500-Year Old Puzzle Behind All the Mysteries of Time and Space; What’s Luck Got to Do with It? The History, Mathematics, and Psychology behind the Gambler’s Illusion; Enlightening Symbols: A Short History of Mathematical Notation and Its Hidden Powers; and Fluke: The Math and Myth of Coincidence. The Clock Mirage: Our Myth of Measured Time is his latest book.


Are you ready to unlock the power of compound interest and take control of your financial future? Understanding how to calculate your investment growth can be a game-changer when it comes to maximizing your wealth. Whether you’re a seasoned investor or just starting out, this knowledge is essential to ensure your money is working as hard as possible for you. In this guide, we will break down the concept of compound interest and show you step-by-step how to calculate your investment growth. From simple interest to compound interest, we’ll explain the differences and why compound interest is the key to exponential growth. We’ll also provide you with practical examples and tools to help you visualize the potential of your investments over time. So, if you’re ready to take charge of your financial destiny and make your money work for you, let’s dive in and discover the incredible power of compound interest.
What is compound interest?
Compound interest is a powerful concept that allows your investments to grow exponentially over time. Unlike simple interest, which only calculates interest on the initial amount invested, compound interest takes into account the accumulated interest as well. This means that the interest earned in each period is added to the principal, and subsequent interest calculations are based on this new total. In other words, compound interest is interest on top of interest. This compounding effect can have a significant impact on the growth of your investments.
Understanding the concept of compound interest is crucial for anyone looking to build long-term wealth. By harnessing the power of compounding, you can potentially turn a small investment into a substantial nest egg over time. The key is to start early and let time work in your favor. Even small contributions made consistently can lead to significant wealth accumulation thanks to the magic of compounding.
The power of compound interest
The true power of compound interest lies in its ability to generate exponential growth. As your investments grow, the amount of interest earned also increases. This creates a compounding effect where your money starts working for you, instead of the other way around. Over time, the growth curve becomes steeper, and your investments can experience rapid growth.
To illustrate this, let’s consider a hypothetical scenario. Imagine you invest $10,000 at an annual interest rate of 5% with monthly compounding. After one year, you would earn $500 in interest. However, instead of withdrawing the interest, you reinvest it, bringing your new principal to $10,500. In the second year, you would earn interest on the new principal, resulting in $525. This process continues, and with each passing year, the interest earned becomes larger. By the end of 10 years, your initial investment would have grown to approximately $16,386. This is the power of compound interest in action.
Understanding the compounding period
The compounding period refers to how often interest is calculated and added to the principal. It can have a significant impact on the growth of your investments. The more frequent the compounding, the more interest you will earn. For example, monthly compounding will generate more interest compared to annual compounding.
To understand the impact of compounding periods, let’s consider two scenarios. In the first scenario, you invest $10,000 at an annual interest rate of 5% with monthly compounding. In the second scenario, you invest the same amount at the same interest rate, but with annual compounding. After 10 years, the investment with monthly compounding would grow to approximately $16,386, while the investment with annual compounding would only reach around $16,289. This shows that even a small difference in compounding frequency can lead to a significant disparity in investment growth.
How to calculate compound interest
Calculating compound interest may seem daunting at first, but it’s actually quite simple once you understand the formula. The formula to calculate compound interest is:
A = P(1 + r/n)^(nt)
– A is the future value of the investment
– P is the principal (initial investment amount)
– r is the interest rate (expressed as a decimal)
– n is the number of compounding periods per year
– t is the number of years the money is invested
Let’s break down the formula using an example. Suppose you invest $5,000 at an annual interest rate of 6% with quarterly compounding. You want to calculate the value of your investment after 5 years. Plugging the values into the formula, we get:
A = 5000(1 + 0.06/4)^(4*5)
Simplifying the equation, we have:
A = 5000(1.015)^20
Using a calculator, we find that the future value of your investment after 5 years would be approximately $6,416.40.
Examples of compound interest calculations
To further illustrate the power of compound interest, let’s consider a few examples.
You invest $1,000 at an annual interest rate of 8% with monthly compounding. After 20 years, your investment would grow to approximately $4,661.39.
You invest $10,000 at an annual interest rate of 3% with quarterly compounding. After 10 years, your investment would grow to approximately $13,439.85.
You invest $500 at an annual interest rate of 6% with annual compounding. After 15 years, your investment would grow to approximately $1,197.96.
These examples demonstrate how compound interest can significantly boost the growth of your investments over time. The longer you stay invested and the higher the interest rate, the greater the impact of compound interest.
The impact of different interest rates and compounding periods
The interest rate and compounding period have a direct impact on the growth of your investments. A higher interest rate or more frequent compounding can accelerate the growth of your money.
To understand this concept, let’s compare two scenarios. In the first scenario, you invest $10,000 at an annual interest rate of 4% with monthly compounding. In the second scenario, you invest the same amount at the same interest rate, but with quarterly compounding. After 20 years, the investment with monthly compounding would grow to approximately $24,117.14, while the investment with quarterly compounding would only reach around $23,847.23. This shows that a higher compounding frequency can result in slightly higher investment growth.
Similarly, if we compare two scenarios with the same compounding frequency but different interest rates, we can observe a significant difference in investment growth. For instance, if you invest $10,000 at an annual interest rate of 3% with monthly compounding, your investment would grow to approximately $18,061.41 after 20 years. However, if you increase the interest rate to 5%, the investment would grow to approximately $26,533.88. This demonstrates the exponential growth potential of compound interest when paired with higher interest rates.
The Rule of 72: A quick way to estimate investment doubling time
The Rule of 72 is a handy rule of thumb that allows you to estimate the time it takes for an investment to double based on the compound interest rate. The formula is simple:
Years to Double = 72 / Interest Rate
For example, if you have an investment with an annual interest rate of 6%, it would take approximately 12 years for your investment to double (72 / 6 = 12). This rule provides a quick and easy way to gauge the potential growth of your investments over time.
Strategies to maximize compound interest growth
Now that you understand the power of compound interest, you may be wondering how to maximize its growth potential. Here are a few strategies to consider:
1. Start early: The earlier you start investing, the longer your money has to compound and grow. Time is one of the most critical factors in generating substantial wealth through compound interest.
2. Increase your contributions: By consistently contributing more money to your investments, you can accelerate the growth of your portfolio. Even small increases in contributions can have a significant impact over time.
3. Take advantage of tax-advantaged accounts: Utilize tax-advantaged accounts such as 401(k)s or IRAs to maximize the growth of your investments. These accounts offer tax benefits that can help your money grow faster.
4. Diversify your investments: Spread your investments across different asset classes to minimize risk and maximize returns. Diversification can help you weather market fluctuations and increase the overall growth potential of your portfolio.
5. Reinvest your earnings: Instead of withdrawing your earnings, reinvest them to take advantage of the compounding effect. By reinvesting your dividends or interest payments, you can accelerate the growth of your investments.
6. Regularly review and adjust your portfolio: Keep an eye on your investments and make necessary adjustments to ensure they align with your financial goals. Regular reviews can help you identify under-performing investments and make informed decisions to optimize your portfolio’s growth.
By implementing these strategies, you can harness the full potential of compound interest and maximize the growth of your investments.
Compound interest vs simple interest: Key differences
While compound interest is a powerful wealth-building tool, it’s essential to understand the key differences between compound interest and simple interest.
Simple interest is calculated based on the principal amount and the interest rate, without taking into account any accumulated interest. Unlike compound interest, simple interest does not have a compounding effect. This means that the interest earned remains constant throughout the investment period.
On the other hand, compound interest considers the accumulated interest and calculates future interest based on the new principal amount. This compounding effect can lead to exponential growth, making compound interest a more favorable option for long-term wealth accumulation.
In summary, compound interest allows your investments to grow exponentially over time, while simple interest only calculates interest based on the initial principal. Understanding these differences is crucial when making investment decisions and maximizing your wealth-building potential.
Understanding how to calculate your investment growth is essential for anyone looking to take control of their financial future. Compound interest is a powerful tool that can turn a small investment into significant wealth over time. By harnessing the compounding effect, your money can work for you, generating exponential growth and securing your financial well-being.
In this guide, we’ve explored the concept of compound interest, its calculation formula, and practical examples. We’ve also discussed the impact of different interest rates and compounding periods, as well as strategies to maximize compound interest growth. By applying this knowledge and taking advantage of the power of compound interest, you can unlock the full potential of your investments and achieve your financial goals.
So, are you ready to take charge of your financial destiny? Start harnessing the incredible power of compound interest today and watch your investments grow beyond your expectations. Remember, time is on your side, and with the right knowledge and strategies, you can unlock the doors to financial freedom.


Carbohydrate counts in gluten free grains vary wildly. The nutritional aspects of a grain, including its protein and fiber contents are not standard fare. If you are living with type 1 diabetes and celiac disease, managing these grains and counting carbohydrates demands daily vigilance.
A tenet of dietary dogma is that fiber isn’t digestible and therefore has no effect on the blood sugar. Because insulin is given in relationship to carbohydrates consumed, it is important to get your math right. By subtracting the appropriate amounts of fiber in foods from the total carbohydrates, you get a “net carb” total.
I crafted a list of common gluten free grains, their carbohydrate load, fiber and net carbohydrate totals associated with them. Take a gander and see how your choices rank.
Dietary recommendations and teaching morph over the years and have landed in a math equation of sorts with current standards. The basic tenet of today’s teachings focus on fiber content/ per serving that exceeds 5 grams. The old school way I learned was to simply subtract the fiber from the carbs and that is your net carb total. I actually still stick with this for myself and my daughter with ease and great success. You will notice the old school math on my chart below for net carb counts.
Current recommendations from the American Diabetes Association
According to current dietary teachings from the American Diabetes Association, “Because fiber is not digested like other carbohydrates, for carbohydrate counting purposes, if a serving of a food contains more than or equal to 5 grams of dietary fiber, you can subtract half the grams of dietary fiber from the total carbohydrate serving of that food.”
Carbohydrate Counting for Dummies (a-b=c if b>or =5)
For example if the carbohydrate count of a serving of food is 30 grams, and the fiber content is 5 grams you would subtract 2.5 grams from the total carbohydrate count. In this instance your total carb count would be [30-2.5= 27.5].
Which ever way you choose to calculate your carbohydrates, it is worth looking at this chart to see the fiber totals associated with common gluten free grains.


Through an immersive, gamified learning experience, Adapted Mind Math helps K-6 students improve their math skills through instruction and exercises. All of Adapted Mind Math’s offerings are in line with common core standards and are designed specific to each grade level.
Learning math does not have to be an exercise in tedium. The Adapted Mind platform is specifically designed to create a fun and interactive learning environment by operating with a game-centric setting, incentivizing learning through virtual prizes.
As students progress through the learning exercises, they earn points, which they can spend on collecting these virtual prizes. They only way they get points, however, is by learning and mastering new skills. By incentivizing the process, students are more likely to push through barriers and reach new levels of learning. According to a survey of parents whose children have taken part in the platform, more than 95 percent of them show an increase in both confidence and overall math ability.
Adapted Mind Math is a provider of innovative online materials that help young learners navigate complex mathematical concepts and boost competency. Subject areas addressed by Adapted Mind Math include dividing mixed numbers. This starts with turning mixed numbers into improper fractions, such that 1 5/6 becomes 11/6.
Keeping the first fraction constant, the second fraction is flipped over and multiplied across instead of divided. The multiplication is undertaken horizontally, above and below the fraction line. For example, take 2 1/2 divided by 1/4. The first fraction is written as 10/4, the second as 4/1. Multiplying out gives the fraction 40/4, which simplifies down to 10.
AdaptedMind provides targeted repetition that benefits those new to dividing mixed numbers. The program defines problem areas based on students’ correct and incorrect answers, and deploys a tailored set of questions.
Bringing together the knowledge provided by mathematics teachers and the creativity of game designers, Adapted Mind Math offers K-6 mathematics students the opportunity to learn while having fun. Addressing a wide range of topics, including algebra and geometry, Adapted Mind Math adapts to the needs of learners to create an interactive and immersive experience.
As students reach middle school, they will be confronted with the various classifications of triangles, each of which must be understood. They are:
Equilateral Triangles. These have three equal sides and angles. In all cases, each individual angle in an equilateral triangle will measure 60 degrees.
Right Angle Triangles. As the name suggests, these are triangles that feature one right angle. The other two angles will be acute, which means they measure less than 90 degrees.
Isosceles Triangles. These are similar to equilateral triangles, except that an isosceles triangle will only have two equal sides and angles.
Scalene Triangles. Perhaps the most awkward classification of triangles, these have no equal sides. Technically, a right angled triangle is also classified as a scalene triangle.
Acute Angle Triangles. These are triangles in which all three angles are below 90 degrees. Both isosceles and equilateral triangles are also acute angle triangles.
Obtuse Angle Triangles. These are triangles in which one angle is above 90 degrees.
An innovative educational platform provider, Adapted Mind Math is focused on creating fun ways for children to learn mathematical and reading skills. Placing the emphasis on adaptive learning, the Adapted Mind Math platform makes use of incentives, such as virtual badges and trophies, to reward users for their progress.
Such incentives have become common in gaming and are used by designers to encourage players to explore their games in more depth. Badges and trophies not only give players something to work toward, but also set higher expectations than players may have placed on themselves.
Some, including Jamie Madigan in an article posted on The Psychology of Games, a website devoted to examining how video games and psychology intersect, argue that the use of such virtual achievements offers a point of reference that players fixate on until they achieve the goals required to unlock the associated badge or trophy.
Further, achieving goals leads to a sense of satisfaction that encourages players to push toward the next achievement so they can experience the same feelings again. This desire to pursue similar goals increases player engagement with the game and leads to them spending more time with it, resulting in increased familiarity and mastery of the topics around which the game is built.
Adapted Mind Math offers multiple ways to work on math concepts. And because there are a number of ways to practice your mathematics skills, Adapted Mind Math’s available resources can be used to explain difficult concepts from a different angle.
Some of the more common ways to practice math is through repetition and worksheets. These types of exercises give you the chance to practice concepts through a variety of different examples and are a mainstay of elementary education. However, until a concept has been fully grasped, practice sheets can end up being more of a frustration than a help for students. Another resource that can be used are dynamic practice sections on websites that adapt based on what questions you answer correctly. This type of resource gives you more practice with concepts that are unfamiliar or more difficult, and they allow you to move past sections that are more easily grasped, resulting in a more focused practice based on the individual needs of each student.
Adapted Mind Math uses web-based instructional videos and gaming elements to improve a child’s math skills. Adapted Mind Math problems cover a range of mathematical topics, from basic addition and subtraction to advanced algebra and geometry.
Math fluency is a term used to describe a child’s ability instantly to recall the basic components of mathematics. Children generally begin learning addition and subtraction in the first grade, and it is around this time that the development of math fluency begins. Though a first grader’s math fluency may be limited to relatively simple concepts such as 0+0, and later 9+8, these instances of recall serve as the building blocks for more complex equations further down the line.
Basic addition facts can be defined as single digit equations with sums lower than 19. The operations used to solve a problem such as 3+2 should become reflexive as a child progresses in his or her education, to the point that teaching a child to add double digit figures with sums exceeding 19 or 20, equations that require the mathematical skill of regrouping, represents a progressive lesson rather than a brand new academic concept.
Of course, the development of math fluency is not as simple as repeatedly presenting children with addition and subtraction facts. Children learn at vastly different rates, regardless of age and intelligence, and a child must first possess the basic concepts of mathematics before initiating fluency. For example, teachers must first convey to students the concept that the order of numbers does not affect an equation. Without an understanding of basic concepts such as this, the development of math fluency will not progress as intended.


OBJECTIVE To identify whether age, sex, or breed is associated with crown height of the left and right maxillary first molar tooth (M1) measured on CT images, to develop a mathematical model to determine age of horses by use of M1 crown height, and to determine the correlation between M1 crown height measured on radiographic and CT images.
SAMPLE CT (n = 735) and radiographic images (35) of the heads of horses.
PROCEDURES Crown height of left and right M1 was digitally measured on axial CT views. Height was measured on a lateral radiographic image when available. Linear regression analysis was used to identify factors associated with crown height. Half the data set was subsequently used to generate a regression model to predict age on the basis of M1 crown height, and the other half was used to validate accuracy of the predictions.
RESULTS M1 crown height decreased with increasing age, but the rate of decrease slowed with increasing age. Height also differed by sex and breed. The model most accurately reflected age of horses < 10 years old, although age was overestimated by a mean of 0.1 years. The correlation between radiographic and CT crown height of M1 was 0.91; the mean for radiographic measurements was 2.5 mm greater than for CT measurements.
CONCLUSIONS AND CLINICAL RELEVANCE M1 crown height can be used to predict age of horses. Results for CT images correlated well with those for radiographic images. Studies are needed to develop a comparable model with results for radiographic images.


Maths Worksheets Level 3
Price: £14.99 (Excluding VAT at 20%)
Age Group: General
Order Number: 13025
A collection of worksheets graded according to ability. It comes in two photocopiable books covering skills and knowledge, one at level 2 and the other at level 3. The worksheets are in an attracive illustrated style that will appeal to Junior pupils. Both books contain 40 worksheets which cover work on numbers, calculations, data and space.


Dr. Moritz Lehmann, Computational Fluid Dynamics software developer FluidX3D, shared a visualization of the solution to one of the most difficult problems of this kind – the calculation of air flows around a supersonic passenger plane Aérospatiale-BAC Concorde on landing.
Computations were performed on the GigaIO SuperNODE GPU server equipped with 32 AMD Instinct MI210 GPUs with 64 GB and a total volume of video memory of 2 TB.
The calculation task is the simulation of air flows for 1 second during the flight of the Aérospatiale-BAC Concorde at a speed of 300 km/h with an angle of attack of 10°. The Reynolds number (a characteristic number in hydrodynamics based on the ratio of the inertia of a gas, liquid or plasma flow to its viscosity) based on the wingspan is 146 million (that’s a lot).
The simulation resolution is 2976 × 8936 × 1489 = 40 billion blocks, with a block size of only 12.4 mm³. 67,268 simulation steps were computed in 29 hours, plus 4 hours spent rendering 5 × 600 frames (five different angles) at 4K resolution, for a total of 33 hours on the GigaIO SuperNODE server. One render frame is based on 475 GB of data, so 600 frames is 285 TB of data.
The entire simulation is a test of the recently added free-slip algorithms at FluidX3D at object boundaries, which allow for a more accurate model for the turbulent air/fluid boundary layer.
According to the author, on the same hardware, some commercial computational fluid dynamics programs, such as Ansys or Star-CCM+, take several years for similar simulations. FluidX3D does it in a few days.
And, well, in general, it is very beautiful and helps us remember why powerful computers are actually needed.
View source version on mezha.media: https://mezha.media/en/2023/08/03/one-of-the-most-difficult-problems-of-computational-fluid-dynamics-concorde-during-landing-video/


As a kid, I was a huge puzzle freak.
I liked jigsaw puzzles; the games Rush Hour, Tantrix and Labyrinth; brainteasers; and books like Rowan of Rin and Artemis Fowl and The Eleventh Hour. I loved it when books gave me a chance to solve the problems with the characters. Often, I’d flip back to check out a problem again.
Once, in Grade Three, I had a brilliant teacher who set us a basic substitution code – A = W, but B = F, C = J etc – and I liked it so much, he set me another one. I spent all of playtime figuring it out. I still remember the message: ‘when I was born, I was so surprised that I couldn’t speak for a year and a half!’
Riddles have a very long history indeed. The oldest riddle ever is probably the riddle of the Rhind Mathematical Papyrus, an Ancient Egyptian problem from 1650 BCE that is similar to the ‘As I was going to St Ives’ riddle:
Seven houses keep seven cats; each cat eats seven mice; each mouse would have eaten seven ears of corn; each ear of corn would have produced seven hekat of grain. How many things are described?
Apparently, there are lots of possible answers. The mathematical answer is 19,607, which makes it a straightforward multiplication problem. But you could also say, for instance, that the answer is five: houses, cats, mice, corn and units of volume are five different things.
Riddles show up a lot in folklore and fairytales, which is probably why they show up so often in fantasy. Not complaining! It gives me the perfect excuse to put riddles in my own stories. Here are a few from my research and development for The Celestial Kris and The Grandest Bookshop in the World. Some of these didn’t end up fitting into the stories. For instance, in The Celestial Kris, the riddling character has a fixation on family, motherhood and being loved, so all of her conundrums (yes, it’s a word, we are descriptivists around here) are kind of in that vein.
Below every riddle is the answer in white text, so highlight the words to see them.


Students from across the city came to the Intrepid on Thursday to take part in a bundle of science, technology, engineering and math or STEM activities as part of a celebration for Time Warner Cable's Connect a Million Minds initiative.
The celebration brought Time Warner Cable executives and local non-profits together to commemorate the initiative reaching a million mind milestone, connecting one million young people to science, technology, engineering and math.
Students got to share their love for STEM and presented the work they have been doing all summer with their organizations.
Executives say while they have reached one million minds, it's not stopping there.
"No question that this is not the end, it’s another milestone along the way, and we've changed our mantra to connecting a million minds and counting. So it keeps going from here," said Time Warner Cable Chairman and CEO Rob Marcus.
Time Warner Cable is NY1's parent company.
To find hands on science, technology, engineering and math opportunities in your community visit connectamillionminds.com.


By Morton Swimmer, Mark Chimley, and Adam Tuaima
Quantum mechanics enables us to build computers that are, in some ways, much more powerful than classical computers. Small, gated quantum computers and experimental quantum computers already exist, but the world might soon see the arrival of quantum computers capable of demonstrating quantum advantage over their classical counterparts. When that happens, it will leave little time for the lengthy process of migrating to new cryptographic algorithms, so it’s best to plan ahead and consider post-quantum cryptography now.
In this entry, the second in our series on post-quantum cryptography, we delve into the history of quantum computing, its foundation in quantum mechanics, and the kind of complex problems quantum computers will be able to solve.
To better comprehend quantum mechanics, it helps to understand how mathematics evolved over time, and with it, our understanding of nature.
Early in history, people’s mathematical understanding of the world was simpler than now: at first, only natural numbers (1, 2, 3, 4, 5, etc.) were used to count. Later, negative numbers (…, -3, -2, -1, 0, 1, 2, 3, ….) were incorporated as well. Eventually, multiplication and division were needed, and these required the concept of fractions created from positive integers. These were classified as rational numbers.
Rational numbers serve well in many practical applications of physics or classical mechanics, where the goal is to measure or approximate something to a sufficient degree of accuracy. However, when the ancient Egyptians wanted to calculate the area of a circle, they tried using fractions in their formula, but the results were not quite accurate. Further examination revealed that a certain number, which will eventually be called pi (symbolized by π and has the value of 3.14159…), can instead be used for this operation. This sparked the idea that there are numbers that cannot be represented by fractions. We call such as real numbers because there’s a tacit assumption that they describe how nature works. So, for a long time, people believed that the world exists on a continuum of scales.
However, the advent of nuclear physics in the late 19th and early 20th centuries started to challenge this view. The smallest unit of matter was initially thought to be the atom, thus it was so named — the ancient Greek “atomos” translates to “indivisible,” or more literally “uncuttable.” In the late 19th century, however, Sir Joseph John Thompson discovered a sub-atomic particle, the electron, and proposed the ‘plum pudding’ model of an atom. In the early 20th century, Ernest Rutherford first theorized the existence of protons and neutrons in an atom’s nucleus, a fact then proven by James Chadwick. Rutherford proposed a model of the atom where the electrons orbited the nucleus like planets around the sun.
Figure 1. Rutherford Atomic Model
But trouble was on the horizon for the planetary model of an atom. Viewed through the lens of classical electromagnetic physics, the electrons would have to continuously lose energy and their orbits would decay rapidly, resulting in the collapse of the atom; but this was not happening.
Meanwhile, Albert Einstein proved Max Planck’s hypothesis that matter has both particle and wave characteristics, for which he later won the Nobel Prize. This is called the wave/particle duality. With this, Niels Bohr, one of the founding fathers of quantum theory, presented a superseding model of the atom in which electrons occupy a set of discrete (or quantized) positions (orbits) based on their energy levels. Describing each orbital by the wave function of the electron explains why the orbitals do not collapse: the smallest orbital is the one composed of a single standing wave. Electrons can move from one discrete state (quantum) to another, but they do so discontinuously; that is, they jump between quanta in a quantum leap. Each jump goes from one wave harmonic to another. The earlier assumption that the world is based on real numbers was proven incorrect. Wolfgang Ernst Pauli later determined that no two electrons can inhabit the same state at the same time, making the model useful in predicting an atom’s chemical properties.
Figure 2. Bohr Model
Most of these might be hard to relate to everyday experiences, but they sit coherently within the quantum mechanical model in terms of mathematics. While many scientists in the early 20th century — for instance, Einstein — were skeptical of quantum mechanics, over time, the field has proven to be sound and useful.
Other observations that fit the quantum model include Heisenberg's uncertainty principle, which states that it is not possible to precisely determine both the momentum and position of an electron. The more accurately we discern one aspect (either momentum or position), the less accurately the other aspect is determined. This is not a limitation of the apparatus. This is a fundamental property of quantum mechanics and plays an important role in quantum computing.
Figure 3. Heisenberg’s Uncertainty Principle
As shown in the left portion of the illustration above, if we use a high frequency to determine the location of an electron, we will not know its momentum with good accuracy. If we use a low frequency, we can find its momentum with good accuracy but will then not know its location well. In the quantum mechanical world, any measurement comes with a trade-off.
Heisenberg’s uncertainty principle changes how we think about computation. While in a classical computer, reading from a memory location doesn’t change it, the same is not true in a quantum computer. Reading the state of a quantum system will invalidate its state, so algorithms need to be designed appropriately.
Two other important topics in quantum mechanics are superposition and entanglement. These are central to quantum computing. Together, they make quantum computing possible.
Until they are measured, particles can exist in infinitely many states. Take an electron, for instance. It is said that it can spin on the x, y, and z axes, as shown in Figure 4. When an electron’s spin is clockwise along some axis, we can call this the “1” state and when it is spinning counterclockwise, we call this the “0” state. However, before measuring the spin, the electron is in a superposition of all possible states. Think of a spinning coin: we cannot say it is in a heads or tails state while it is spinning. Rather, it is in some other in-between state. It’s in a superposition. (Note that the electron doesn’t actually spin. This is just a measurable property of it that we chose to call “spin”.)
The particle is in superposition while it is not observed. When the spin is measured, the state will “collapse” on one choice dictated by its probability distribution, just like when the coin flops down on one side or the other with a 50% chance of being either.
Figure 4. An electron can “spin” on the x, y, and z axes
This is the basis of the famous Schrödinger’s Cat thought experiment (Figure 5). An imaginary cat is placed in a box with a vial of poison and a radioactive source. If radioactive decay is detected, the vial of poison breaks and ends the imaginary cat’s life. But radioactive decay is a random process which may or may not happen.
While this was originally meant to demonstrate the absurdity of quantum mechanics, it turned out to be a reasonable explainer of superposition. When the box is shut, there is no way of observing what happens inside. Whether the cat is alive or dead cannot be concluded. From an observer’s point of view, the cat is in a superposition of dead and alive. In quantum mechanics, opening the box causes the superposition to collapse into a single state of the cat being either dead or alive.
Figure 5. The Schrödinger’s Cat thought experiment
In a classical computer, the smallest unit of storage is called a bit, which can either take a 0 or a 1. In a quantum computer, particles can be used to perform calculations, and these are called qubits (quantum bits). Qubits are different from ordinary bits in that through superposition, they can assume an infinite number of states at the same time, which gives them exponentially more computational power. Qubits are used in quantum circuits, which are comparable to algorithms on a classical computer. The goal of these circuits is to run to the end and amplify the correct result so that the answers can be read by measuring the resulting states. In practice, the circuit is re-run thousands of times to obtain a probability distribution from which the answer is derived. However, a circuit needs gates, and the use of multiple gates requires entanglement. The following section delves into this concept.
In a classical computer, a logic gate can take bits as input and produce a value as output. For example, a “1” bit and another “1” bit combined at an AND gate will result in “1”. On their own, these bits do nothing. However, in a quantum computer, two particles can become “entangled,” which means an operation on one will influence the state of the other, even if they are far apart. This is like picturing two spinning coins that always land down the same way: either showing both heads or both tails. The two particles are not communicating with each other to do this. Instead, imagine them as twins that behave the same even when they are separated. In this way, they are not violating special relativity that disallows travel at the speed of light or faster. In the macro world, however, this would be strange, and we would not expect coins to behave this way.
Einstein was not a fan of the idea and described it as “spooky action at a distance.” It might take some convincing to believe that that two particles could influence each other remotely, but it is mathematically proven by and shown in experiments. Between 1980 and 1982, Alain Aspect performed various experiments proving entanglement is real, for which he won the Nobel Prize in 2022.
Entanglement makes useful algorithms possible. By linking two qubits, conditional quantum gates can be created in quantum circuits. With that, algorithms can be implemented.
When the first serious efforts to build a quantum computer were made in the mid-1990s, one of the main questions raised was, what would it be good for? Actually, this very question serves as the answer itself to one of the main motivations why these computers are developed: they were built so their full potential can be discovered and unleashed.
The original idea of quantum computing was proposed by Paul Benioff and Richard P. Feynman around 1980. Simulations of physical systems, for instance tracking all interactions between atoms in a molecule, are often impossible to run on classical computers because the difficulty of the problem grows exponentially with the number of atoms. In contrast, quantum computers can tackle exponentially larger problems for every qubit in the quantum computer and can deal with this complexity.
Now, while experimental quantum computers are emerging, there are still challenges on how their usefulness can be most effectively maximized. Machines with sufficiently large numbers of usable qubits remain elusive. Furthermore, scalability and noise management remain a concern. Noise could limit scalability. In this context, noise pertains to electromagnetic radiation, magnetic fields, cosmic rays, and the like. For this reason, most quantum computers are operated at temperatures close to absolute zero (which is 0 Kelvin, -273.15℃, or -459.67 ℉) and are heavily shielded from radiation.
With the current crop of quantum computers, attempts to mitigate noise are done by using redundant noisy physical qubits to create noise-free logical qubits. The current generation of quantum computers is called Noisy Intermediate-Scale Quantum (NISQ) computers. The number of logical qubits is much lower than the number of physical qubits in these systems. This means that a useful quantum computer will have to have several dozen times as many physical qubits as required logical qubits in order to execute a given quantum circuit.
Despite the challenges, many algorithms for quantum computers have been proposed, showing how versatile the platform may one day be. As of now that question of their purpose still stands, although perhaps not much longer.
Quantum advantage pertains to the milestone where a quantum computer can solve problems faster than a classical computer can. While several claims of quantum advantage have been made recently, none of them solves useful problems. They are therefore not convincing — yet. Because the simulation of physical systems often requires far fewer qubits, we will probably see the first practical uses here. On the other hand, a quantum computer for solving large computer science problems is probably over a decade away. These simulations would be valuable as they would enable major chemical, physical, and medicinal discoveries. This is why the development of quantum computers has garnered so much attention lately.
In an attempt to horizontally scale the current generation of experimental quantum computers, there have been experiments using circuit knitting or surface-code architectures. These experiments take small quantum computers and combine them using classical or quantum connections to form a larger machine. However, it is not clear if quantum advantage can be achieved this way, as these separate circuits need to be connected, which incurs overhead and a change to the algorithms. The last limiting factor is the number of quantum gates that can be used, but little has been published on this for existing quantum computers.
Small, gated quantum computers containing a few hundred physical qubits, are now available as a cloud service from various vendors. Adiabatic quantum computers (AQC) that are restricted to a special optimization process already have thousands of qubits but have yet to show meaningful speedup over classical computers.
A recent finding from a university in Switzerland showed that merely quadratic speedup over classical algorithms will not be enough to compensate for the overhead that quantum computers have in loading the data and circuit. Given this finding, there are some doubts that they will ever be useful for big-data applications at all, unless some engineering solution is found.
As mentioned, quantum computers can solve some problems in minutes that would take years on a classical computer. In this section, let’s look at the classes of problems that quantum computers can solve.
A complexity class refers to a set of problems that are solvable with relatable time requirements on a Turing machine, which is a theoretical computational model. Without going into details, a nondeterministic Turing machine is exponentially faster than the deterministic variant. We strive to use polynomial-time algorithms (or better) because the time they take only goes up moderately with the problem size. We call these “P” problems. Many real-world problems are “NP” problems that require much greater resources and typically grow in time and space requirements exponentially with the size of the input problem. We say that a problem is NP-Complete if it can be proven that no algorithm solves the problem better than a non-deterministic Turing machine would need in polynomial time. The entire space of decidable problems that a Turing machine can solve is called PSPACE.
The complexity class for quantum algorithms solvable in polynomial time is called Bounded-Error Quantum Polynomial Time complexity (BQP). This class is not thoroughly explored, but we believe it includes the whole P class and some of the NP and PSPACE classes, but no NP-Complete problems (as shown in Figure 6 below). Solving some of the NP problems in BQP time complexity is a game changer, even if it might disappoint some that NP-complete problems still stay exponentially difficult.
Figure 6. Quantum computer complexity classes
What quantum computers cannot do is solve completely intractable and (classically) undecidable problems. And while there are problems that are difficult for classical computers but relatively easy for quantum computers, the reverse is also true.
The rise in interest in quantum computing algorithms has also revived creativity in classical algorithm design. Some heuristic algorithms can now be seen emerging from this research. These are often optimization algorithms that are called Quantum Inspired Optimization (QIO) or sometimes Quantum-Inspired Algorithms (QIA). Therefore, research into quantum computing is leading to speedups in classical real-world algorithms, which might turn out to be more of a game changer than quantum computers themselves.
Quantum computers are vastly different from classical computers in fundamental ways. This is shown through a variety of factors, from the probabilistic nature of their operation to the resulting complexity classes of the algorithms that can be implemented on quantum computers. This article looked into the history of quantum computing, pertinent concepts in quantum mechanics, and the type of problems quantum computers can help solve. It also touched on the types of gates that can be used in quantum circuits. In the next parts of this research, we will look at how quantum computer cryptanalysis works and why there are worries about these nascent computers.
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Math Arcade V1.2
Put away your calculator and give Math Arcade a try! Whether you’re a grade school student learning basic math for the first time or an old timer in need of a refresh, Math Arcade will hone your mental math skills like nothing else!
Math Arcade is a collection of 8 different games with 5 skill levels each. The games vary from simulated flash card learning for addition, subtraction, and multiplication, to full blown long hand addition, subtraction, multiplication, and division. There is even a mind-bending equation builder that will be sure to test even the best mental math wizards!
There is both a practice and an arcade mode for each game. In practice mode, you can try as many problems as you want. In arcade mode, up to 4 people at once can compete solving a fixed number of problems in a race against the clock and accuracy for a shot at the high score table.
At the end of each arcade game attempt, a summary sheet is provided to review your performance. There is even printing support so you can have a permanent record of any summary report.
The demo version of Math Arcade has 2 of the 8 games enabled for your evaluation. If you enjoy this program, please consider purchasing the full version to help support independent software development.
System Requirements: Windows OS, 800×600+ desktop resolution, 300+ MHz CPU recommended
Download the demo version here: MathArcadeDemo.zip (1.04 MB)
Order the full version of Math Arcade V1.2 for $9.95


During 2020, despite the challenges posed by a lack of face to face tuition, three of our Grade 10 learners performed exceptionally well in the Advanced Programme in Maths (AP Maths). These learners namely: Husna Haffeejee, Noormohamed and Moosa Abdulla, all achieved a mark above 90% for the year in the exam offered by Advantage Maths. All lessons were conducted virtually. Well done to all these learners. We hope they will pursue with AP Maths and continue to achieve excellent results.


Chennai Mathematical Institute: Admission 2004
The Chennai Mathematical Institute invites applications for the following programmes:
· BSc (Honours) in Mathematics and Computer Science (3 year integrated course).
· BSc (Honours) in Physics (3 year course)
· MSc in Mathematics (2 year course)
· MSc in Computer Science (2 year course)
· PhD in Mathematics
· PhD in Computer Science
About the courses
The courses are conducted by the Institute with the cooperation of the Institute of Mathematical Sciences, Chennai, and the School of Mathematics, TIFR, Mumbai.
The BSc and MSc degrees are awarded by Madhya Pradesh Bhoj (Open) University. The PhD degrees are awarded by University of Madras and BITS, Pilani.
· BSc: 12th standard or equivalent.
· MSc (Math): BSc (Math)/B Math/B Stat/B Tech
· MSc (Computer Science): BE/B Tech/BSc (Computer Science) or BSc (Math) with a strong background in Computer Science
· PhD (Math): BE/B Tech/BSc (Math)/MSc (Math)
· PhD (Computer Science): BE/B Tech/MSc(Computer Science)/MCA
For all the programmes, applicants shortlisted based on their scholastic record will be required to take an entrance examination to be held in Allahabad, Bangalore, Bhopal, Calicut, Chandigarh, Chennai, Hyderabad, Kolkata, Mumbai and New Delhi, on Monday, May 31, 2004. In addition, selection for PhD will involve an interview at Chennai.
How to apply
To obtain the application forms and information brochures, send a DD for Rs 250 in favour of the Chennai Mathematical Institute, payable at Chennai, to the Chennai Mathematical Institute, 92 G N Chetty Road, Chennai 600 017.
Web site: www.cmi.ac.in
The last date for collecting application forms is April 12, 2004.
The last date for submitting completed application forms is April 30, 2004.


A rectangular prism, also known as a rectangular cuboid, has 12 edges. These edges are formed by the intersection of the rectangular faces of the prism.
Rectangular prisms are common three-dimensional shapes encountered in everyday life and mathematics. Understanding their properties and calculations is fundamental in various fields, from construction to mathematics.
Table of Contents
Properties of Rectangular Prism
Rectangular prisms possess distinctive properties that make them useful in diverse applications. These include volume, surface area, and diagonals.
- Volume: The volume of a rectangular prism refers to the amount of space it occupies. It is calculated by multiplying the length, width, and height of the prism.
- Surface Area: Surface area represents the total area covered by all the faces of the rectangular prism. It includes the area of each individual face.
- Diagonals: Diagonals in a rectangular prism are lines connecting opposite vertices, passing through the center of the prism. Understanding diagonal lengths is crucial in various geometric calculations.
How to Calculate Volume of a Rectangular Prism
Calculating the volume of a rectangular prism involves a straightforward formula:
Volume = Length × Width × Height
For example, if a rectangular prism has a length of 5 units, width of 3 units, and height of 4 units, its volume would be:
Volume = 5 × 3 × 4 = 60 cubic units
How to Calculate Surface Area of a Rectangular Prism
Determining the surface area of a rectangular prism requires summing up the areas of all its faces. The formula for surface area is:
Surface Area = 2lw + 2lh + 2wh
Where l, w, and h represent the length, width, and height of the prism respectively.
Diagonals of a Rectangular Prism
The diagonals of a rectangular prism connect opposite vertices, forming lines within the prism. These diagonals are essential in various geometric calculations, such as determining internal angles and spatial relationships.
Real-world Applications of Rectangular Prisms
Rectangular prisms find extensive use in real-world scenarios, including:
- Architecture: Rectangular prisms serve as building blocks for architectural structures, such as buildings and bridges.
- Packaging: Many packaging materials, like boxes and cartons, are designed in the shape of rectangular prisms for efficient storage and transportation.
- Engineering: In engineering projects, rectangular prisms are utilized in constructing components, machinery, and infrastructure.
Importance of Rectangular Prisms in Mathematics
In mathematics, rectangular prisms hold significant importance:
- Geometry: They are fundamental shapes studied in geometry, helping to understand concepts like volume, surface area, and spatial relationships.
- Algebraic applications: Rectangular prisms are often used as models in algebraic equations, facilitating problem-solving and visualization of mathematical concepts.
FAQs on Rectangular prisms
- What is a rectangular prism?
- A rectangular prism is a three-dimensional shape with six rectangular faces, where each face meets at right angles.
- How do you find the volume of a rectangular prism?
- The volume of a rectangular prism is calculated by multiplying its length, width, and height.
- Why are rectangular prisms important in mathematics?
- Rectangular prisms are significant in mathematics due to their relevance in geometry, algebra, and real-world applications.
- What are some real-world examples of rectangular prisms?
- Examples include buildings, boxes, bookshelves, and shipping containers.
- How are diagonals useful in rectangular prisms?
- Diagonals help determine internal angles, spatial relationships, and other geometric properties of rectangular prisms.
Rectangular prisms are versatile geometric shapes with essential properties and applications in various fields. Understanding their characteristics and calculations is crucial for both practical and theoretical purposes.


Up to this point we have worked with unsigned integers, i.e. whole numbers which are zero or bigger than zero. Much like it happened with endianness, the interpretation of the value of several bytes as a negative number depends on the specific interpretation.
In computing there are two main ways to interpret the values of a group of bytes as a negative number: one’s complement and two’s complement.
At the moment GNU poke supports the two complement interpretation, which is really ubiquitous and is the negative encoding used by the vast majority of modern computers and operating systems.
We may consider adding support for one’s complement in the future, but only if there are real needs that would justify the effort (which wouldn’t be a small one ;)).
Unsigned values are never negative. For example:
(poke) 0UB - 1UB 0xffUB
Instead of getting a -1, we get the result of an unsigned underflow, which is the biggest possible value for an unsigned integer of size 8 bits: 0xff.
When using type specifiers like
uint<16> in a
map, we get unsigned values such as 0UB. It follows that we need
other type specifiers to map signed values. These look like
For example, let’s map a signed 16-bit value from foo.o:
(poke) .set obase 10 (poke) int<16> @ 0#B 28515H
Note how the suffix of the value is now
H and not
This means that the value is signed! But in this case it is still
positive, so let’s try to get an actual negative value:
(poke) var h = int<16> @ 0#B (poke) h - h - 1H -1H
Adding two signed integers gives you a signed integer:
(poke) 1 + 2 3
Likewise, adding two unsigned integers results in an unsigned integer:
(poke) 1U + 2U 3U
But, what happens if we mix signed and unsigned values in an expression? Is the result signed, or unsigned? Let’s find out:
(poke) 1U + 2 3U
Looks like combining an unsigned value with a signed value gives us an unsigned value. This actually applies to all the operators that work on integer values: multiplication, division, exponentiation, etc.
What actually happens is that the signed operand is converted to an unsigned value before executing the expression. You can also convert signed values into unsigned values (and vice-versa) using cast constructions:
(poke) 2 as uint<32> 2U
Therefore, the expression
1U + 2 is equivalent to
1U + 2
(poke) 1U + 2 as uint<32> 3U
You may be wondering: why not doing it the other way around? Why not converting the unsigned operand into a signed value and then operate? The reason is that, given an integer of some particular size, the positive range that you can store in it is bigger when interpreted as an unsigned integer than when interpreted as a signed integer. Therefore, converting signed into unsigned before operating reduces the risk of positive overflow. This of course assumes that we, as users, will be working with positive numbers more often than with negative numbers, but that is a reasonable assumption to do, as it is often the case!


Probability and other paradoxes
|Door 2||Door 3||Result|
3) By opening his door, Monty is actually saying to the player, "There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. Your choice of door 1 has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating Door 2, I have shown you that the probability that Door 3 hides the prize is 2 in 3."
4) (Mine) Choosing a door, you know well that this door has 1/3 chances of hiding the car. The remaining 2/3 chances are with the other two doors. Now, after a door is opened by the host, these 2/3 chances don't change, i.e. the third door has still 2/3 chances of hiding the car.
You can play/test the game Let's Make a Deal here (as long as the page is still alive!). You will find out that:
1) If you always stick to the same door, 'win' and 'lose' results are about the same.
2) If you always switch to the other door, 'win' results are much higher than 'lose'!
I continue with some variations of the Monty Hall problem.
There are 3 playing cards on the table, face-down. One of them is Queen of Hearts (QoH). The game "host" asks you to guess which one is it. OK, so you point card C. Then card B is turned over by the "host" and it isn't a QoH, but a non-QoH). So the situation schematically is the following:
|Card A||Card B||Card C|
At this point, you are asked whether you want to switch to card A or not.
You may think, "I don't have any reason to switch cards became there's a QoH and a non-QoH remaining, so the probability of winning is the same by switching on not, it's 50%."
There are two variations. Let's see what happens in each one.
The "host" knows where the QoH card is. And you know that. In this case, the chance of winning is doubled when you switch to the other card (A) rather than sticking with your original choice (card C), because the "host" turned deliberately an non-QoH card over.
There are 3 possible situations corresponding to your initial choice, each with equal probability (1/3):
1) You originally picked non-QoH #1. The "host" has deliberately shown you non-QoH #2.
2) You originally picked non-QoH #2. The "host" has deliberately shown you non-QoH #1.
3) You originally picked the QoH card. The "host" has shown you either of the two non-QoH cards.
If you choose to switch, you win in the first two cases. If you choose to stay with the initial choice, you will win only in third case. So in 2 out of 3 equally likely cases switching wins, since the odds of winning by switching are 2/3. In other words, a player who has a policy of always switching will win on average two times out of the three.
The "host" doesn’t really know where the QoH card is and you know that. In this case, when the "host" turns the non-QoH card over, the probability that you originally picked the QoH card increases from 1/3 to 1/2. And the odds in this case are equal, whether you change your initial choice or not.
Also known as the "exchange paradox".
You are given two indistinguishable envelopes and you are informed that one of them contains twice as much money as the other. You may select any one of the envelopes at random and you can receive the money it is in it. Now, after you select one of them and before opening it, you are given the opportunity to take the other envelope instead. Should you switch to the other envelope?
You may think, "I don't have any reason to take the other envelope since the chances that it is the one with more money are the same with the chances of keeping the one I have already selected."
Suppose that the amount of money in the envelope you first choose is x. Then the other envelope has 50% chances of containing 2x and 50% chances of containing x/2, i.e. a total of 0.5*2x + 0.5*(x/2) = x + 0.25x = 1.25x. So you should switch.
This, however, creates the following paradox: If one switches, by the same reasoning one has to switch back! And this can go on in an endless cycle! So there must be something wrong with above reasoning. Indeed, two things are wrong:
1) To mix different instances of a variable in the same formula like this is said to be illegitimate, so the above reasoning is incorrect.
2) The expectation of 1.25x holds for either of the envelopes!
A box contains three coins. One is normal, one has two heads and the other has two tails. One con falls out of the box on the floor, heads up. What is the probability that the other side is also a head?
(Think of the solution as long as needed before going on . . .)
The table below contains all possible cases:
|Coin||Side up||Side down|
|2 heads||head #1||head #2|
|2 heads||head #2||head #1|
|2 tails||tail #1||tail #2|
|2 tails||tail #2||tail #1|
In the 3 of the above possible coin cases where you get head up, you have 2 cases of head down and one of tail down. So the chances are 2/3.
(I have simplified the description.)
There are three boxes, each with two drawers. Each drawer contains a coin. One box has a gold coin in each drawer (GG), one box has a silver coin in each drawer (SS) and the other has a gold coin in one drawer and a silver coin in the other (GS). A box is chosen at random, and then one of its drawers is opened. The coin in that drawer is gold. What is the chance that the other drawer also contains a gold coin?
Apparently, it seems that the probability is 50%. However, it isn't! Here's why:
Since a gold coin is found, it means that the box is either GG or GS. These 2 boxes have a total of 3 gold coins and 1 silver coin. If we subtract the gold coin already found, there remain 2 gold coins and 1 silver coin. Hence, the probability that the other coin in the box is also gold is 2/3.
Or, we can think in the following way, examining all three, equally possible cases:
1) Drawer G of GS was chosen, so the other drawer contains a silver coin (1/3)
2) Drawer G1 of GG was chosen, so the other drawer contains also a gold coin (1/3)
3) Drawer G2 of GG was chosen, so the other drawer contains also a gold coin (1/3)
In 2 of the three cases the other coin is also gold, i.e. the probability of this happening is 2/3.
Since gold coins are difficult to get and also for getting rid of the boxes, 3 pairs of regular coins may be used, placed on the table and covered, with heads up and tails up as follows: HH, TT and HT. Or 3 pairs of black and red playing cards face down, etc.
You pay a fixed fee to enter a game of chance in a casino, in which a fair coin is tossed repeatedly until a tail appears, ending the game. The pot starts at 2 dollars and is doubled every time a head appears. You win whatever is in the pot after the game ends. Thus you win 2 dollars if a tail appears on the first toss, 4 dollars if a head appears on the second toss and a tail on the second, 8 dollars if a head appears on the first two tosses and a tail on the third, and so on. In short, you win 2^k dollars, if the coin is tossed k times until the first tail appears. What would be a fair price to pay the casino for entering the game?
Let's see what's the average payout: with probability 1/2, the player wins 2 dollars, with probability 1/4 the player wins 4 dollars, with probability 1/8 the player wins 8 dollars, and so on. The expected value is thus EV = 1/2*2 + 1/4*4 + 1/8*8 ... = 1 + 1 + 1 + ... = infinite.
Assuming that the game can continue as long as the coin toss results in heads and that the casino has unlimited resources, this sum grows without bound and so the expected win for repeated play is an infinite amount of money.
Considering nothing but the expected value of the net change in one's monetary wealth, one should therefore play the game at any price if offered the opportunity. Yet, in published descriptions of the game, many people expressed disbelief in the result. It has been found that "few people would pay even $25 to enter such a game" and most commentators would agree. The paradox is the discrepancy between what people seem willing to pay to enter the game and the infinite expected value.
I created and ran a simple computer program to find out what is statistically the average earnings after a large number of rounds. First of all, the average number of tosses, until a tail comes up is found to be 2. Of course, since the first toss is "free", i.e. it doesn't count, since one earns 2 dollars independently of the outcome of the first "toss". Then the 2nd toss has 50% chances to be heads or tails. So each round lasts 2 tosses in average.
Now, here's another paradox within St. Petersburg paradox that I discovered: Since the average of tosses is 2, one would logically expect that the average earnings would be 2^2 = 4 dollars. Although this sounds quite logical, it is very far from the truth. Why?
If the amount of dollars earned after each test were proportional, i.e. 2 dollars were added on each toss, then the average earnings would be indeed 4 dollars. But the amount of earnings does not grow proportionallly but exponentially. Indeed, in the above test, I found that the average earings range from 10 to 25 dollars! So the paradox lies in the big discrepancy between the average number of tosses and the average earnings.
Which of the averages then should be taken as a basis to decide on the amount of the entry fee one would be willing to pay for this chance game? Evidenttly, it's the average earings, which is more real as a statistic and it's what finally counts.Here is the code of the program I used for testing:
DEFINT A -Z DIM tottosses AS LONG, earnings AS LONG, totearnings AS LONG RANDOMIZE TIMER tottosses = 0: totearnings = 0: rounds = 32000 FOR i = 1 TO rounds tosses = 0: earnings = 2 DO tosses = tosses + 1: a = INT(2 * RND) + 1 IF a = 1 THEN earnings = earnings * 2 ELSE EXIT DO LOOP tottosses = tottosses + tosses totearnings = totearnings + earnings NEXT i avtosses = CINT(tottosses / rounds): avearnings = CINT(totearnings / rounds) PRINT 'Average number of tosses, average earnings:'; avtosses; avearnings END
import random random.seed() tottosses = 0; totearnings = 0; rounds = 32000 for i in range(rounds): tosses = 0; earnings = 2 while True: tosses += 1; a = random.randint(1, 2) if a == 1: earnings *= 2 else: break tottosses += tosses; totearnings += earnings avtosses = round(float(tottosses)/float(rounds), 0) avearnings = round(float(totearnings)/float(rounds), 0) print 'Average number of tosses, average earnings:', avtosses, avearnings
If you run it a lot of times, you will see a stable average number of tosses of 2 and an average of earnings, normally lying bewteen 15 and 25 but with big variations each now and then. Don't be surprised if you suddenly see a value that looks like 540!
It is a variation of the "Two envelopes problem" described above.
Two men are given a necktie by their wives as a Christmas present. Over drinks they start arguing over who has the more expensive necktie. They agree to have a wager over it. They will consult their wives and find out which necktie is more expensive. The terms of the bet are that the man with the more expensive necktie has to give it to the other as the prize.
One of the men reasons as follows: Winning and losing are equally likely. If I lose, I will lose the value of my necktie. But if I win, I win more than the value of my necktie. Therefore the wager is to my advantage.
Assuming that both men think equally logically, the other man will think the same thing. So, paradoxically, it seems that both think they have an advantage in this wager. Which is impossible.
What's the flaw in this?
What we call the "value of my necktie" in the losing scenario is the same amount as what we call "more than the value of my necktie" in the winning scenario. Accordingly, neither man has the advantage in the wager. E.g. If the price of one tie is $10 and the other $20, in the losing scenario one loses $20, which is the same amount he wins in the winning scenario.
The Jones family have two children. The older child is a girl. What is the probability that both children are girls?
(Think of the solution as long as needed before going on . . .)
Only two possible cases meet the criteria specified in the question: GB and GG. Since both of them are equally likely, and only one of the two, GG, includes two girls, the probability that the younger child is also a girl is 1/2.
The Jones family have two children. At least one of them is a boy. What is the probability that both children are boys?
(Think of the solution as long as needed before going on . . .)
The possible cases are: GB, BG and BB. Schematically:
|Older child||Younger child|
From the above we can see that the probability that both are boys is 1/3.
A postman knocks on the door of a house where two children live, and the door is answered by a little girl. What are the chances that the other child is also a girl?
How many people do you think it would take to survey, on average, to find two people who share the same birthday? Due to probability, sometimes an event is more likely to occur than we believe it to. In this case, if you survey a random group of just 23 people there is actually about a 50% chance that two of them will have the same birthday. This is known as the birthday paradox. Don't believe it's true? You can test it and see mathematical probability in action!
The birthday paradox, also known as the birthday problem, states that in a random group of 23 people, there is about a 50% chance that two people have the same birthday. Is this really true? There are multiple reasons why this seems like a paradox. One is that when in a room with 22 other people, if a person compares his or her birthday with the birthdays of the other people it would make for only 22 comparisons — only 22 chances for people to share the same birthday.
But when all 23 birthdays are compared against each other, it makes for much more than 22 comparisons. How much more? Well, the first person has 22 comparisons to make, but the second person was already compared to the first person, so there are only 21 comparisons to make. The third person then has 20 comparisons, the fourth person has 19 and so on. If you add up all possible comparisons (22 + 21 + 20 + 19 + … +1) the sum is 253 comparisons, or combinations. Consequently, each group of 23 people involves 253 comparisons, or 253 chances for matching birthdays.
Here is a case that I am sure a lot of people have been confused about or at least thought as a kind of paradox.
We toss a fair coin and get heads up. We toss again and get heads up. As this goes on, everytime we feel more and more strongly that its time for tails to show up. That is, there are more and more chances each time for tails to show up. We have reached a large number of tosses, and we are almost certain that the next toss will be tails up.
Coin tossing may be replaced with drawing a card from a well shuffled deck and getting a red card for a number of consecutive drawings, betting on Red in a roulette, etc.
That at it's time for tails to show up and the accumulated frustration of this not happening is a real feeling we get. We think, "What is the probability that heads continue to show up after, e.g. 9 consecutive times of heads up?" Apparently very low, right? "That would be too much", we could think.
But let's cool down and think in pure probability terms.
There is a basic, fundamental law that applies at any time we are tossing a coin, drawing a card from the deck, betting on Red in a roulette, etc. The probability for heads or a red to come up is 1/2. Similarly of course for tails up or a black card. Always. At any time. Independenty of what has happened until then. The Universe does not keep records. We do. And these records help us producing statistics, etc.
Now, let's look the tossing problem from another view. We could do an experiment and say, "We are going to toss a coin 10 times. What is the probability all outcomes to be heads up?" Easy: It's 1/(2^10), i.e. 1/1024 = ~0.1%. And what is the probability to get at least one tails up? It's 9 times greater, i. ~0.9%. So we note down each time the outcome of a tossing. Then we may think that since at least one tails up is 9 times more probable than 10 heads up, it's 9 times more probable that the 10th toss is tails up. And we may believe this strongly.
Well, if we come back to the fundamental law described above, we had to come down to earth and accept that the probability of a 10th heads up is the same with the probability of a 10th tails up, i.e. 50%. That is, getting HHHHHHHHHH is the same as HHHHHHHHHT.
OK, but what about the 9:1 probability as a statistic? Does HHHHHHHHHH invalidates it? No, because it's an extreme case and statistical formulas always provide for deviations from normal. If we were to repeat the experiment foa a large number of times (e.g. 100 times), we would see that on average outcomes are always as expected. And We could also confirm the 9:1 case.
(It is also known as "cocodile paradox". Although its description involves a totally unrealistic situation, as you will see, and it can be described in a hundred of more realistic ways, I use the original version, as it appears in Wikipedia.)
The Crocodile dilemma is an unsolvable problem in logic. The premise states that a crocodile who has stolen a child promises the father that his son will be returned if and only if he can correctly predict what the crocodile will do.
The outcome is logically smooth (but unpredictable) if the father guesses that the child will be returned, but a dilemma arises for the crocodile if he guesses that the child will not be returned: 1)If the crocodile decides to keep the child, he will violate his terms: the father's prediction has been validated, and the child should be returned. 2) If the crocodile decides to give back the child, he still violates his terms, even if this decision is based on the previous result: the father's prediction has been falsified, and the child should not be returned.
Therefore, the question of what the crocodile should do is paradoxical, and there is no justifiable solution.
Relation to "The king and the jester" riddle
The king, who was tired of his jester and looked for an excuse to get rid of him, calls him in one day and says to him: "Say something, anything you want. If what you say is a lie I will hang you and if it is true I will slaughter you."
The jester thought for a while and then he said something to the king. And lived!
What did he say?
(Hover the mouse pointer over here to see the solution.)
Suppose there is a card with statements printed on both sides:
Front: The statement on the other side of this card is TRUE.
Back: The statement on the other side of this card is FALSE.
Trying to assign a truth value to either of them leads to a paradox.
1. Epimenides was a Cretan who made an immortal statement: "All Cretans are liars"
2. "This sentence is false"
3. "I am lying"
4. What happens if Pinocchio says that his nose is about to grow, knowing that it actually grows ONLY when he is telling a lie?
|A painting by the Belgian René Magritte, featuring a smoking pipe and a message at the bottom saying "Ceci n'est pas une pipe." ("This is not a pipe.") Indeed, the painting is not a pipe, but rather an image of a pipe!|
Socrates' famous phrase: "I know one thing, that I know nothing"
At start of the week, a teacher announces to his students that he will give a test this week.
Now, the students think that locically the test cannot be given on Friday because it would be expected since it is the last remaining day. So the teacher would not give the test on Friday. In the same way, it could not give it on Thursday, because it would be expected since it is the last remaining day, excluding of course Friday. And so on ... So, the teacher cannot give an unexpected test!
Wrong! The teacher may well give the test on Friday, and it will be unexpected, since the students, based on the above reasoning, would not expect it on Friday!
Actually, he can give the test any day, since, based on the above reasoning, the students believe it cannot be unexpected anyway!
Consider a heap of sand from which grains are individually removed. One might well do the following thinking: 1) 1,000,000 grains of sand is a heap of sand. 2) A heap of sand minus one grain is still a heap. 3) Repeated applications of (2) (each time starting with one less grain), eventually forces one to accept the conclusion that a heap may be composed of just one grain of sand.
(From Smullyan's Puzzle Book)
This is a very old problem in logic stemming from ancient Greece. It is said that the famous sophist Protagoras would accept a young person Euathlus as pupil on credit and that hw will get paid for his instruction after he had won his first case. So, Protagoras waited until Euathlus take on a client, but this never happened. So he decided to sue Euathlus for the amount owed.
Protagoras argued that if he won the case he would be paid his money. If Euathlus won the case, Protagoras would still be paid according to the original contract, because Euathlus would have won his first case.
Euathlus, however, claimed that if he won then by the court's decision he would not have to pay Protagoras. If on the other hand Protagoras won then Euathlus would still not have won a case and therefore not be obliged to pay. The question is: which of the two men is in the right?
Answer: The court decision will be based on what has aleready happened, not on what will happen after the court decision is made. So, actually, there's no case at all, since Euathlus has not yet tried to win a case!
Now, here's another thing that Protagoras could do: He should let Euathlus win the case and then sue him again, in which case he should legally get his money since Euathlus would have won his first case!
(This seems a better and more useful math problem. But then, it wouldn't be a paradox!
This is something I have thought in 2004 and it can be seen as simply a problem (successful or defective) method or a paradox.
You bet one chip on Red. If Red comes out, you keep the earned chip and you bet another chip, always on Red. If Black comes out, you bet 2 chips on Red in the next round. If Red comes out, you get 4 chips back, so have earned one more chip (after subtracting the 3 chips you have bet in total). You keep the earned chip and you bet another chip, always on Red. In short, the method can be summarized as follows: "Whenever you win, you keep the earned chip and whenever you loose you double the amount of the previous bet." In this way, you always earn one chip!
Can this be possible? Can this method work on a standard basis?
I have test this by playing roulette games (online and offline) and also with a program I created that produced statistics. Well, After about 5 minutes (i.e. about 100 roulette rounds), I always was over my initial aount (capital). (To this, one has to add the extra 1/37 changes in each round in which both Red and Black lose. But it's too small a probability to affect the results.) So, it looks like it's successful method, insn't it?
However! (There's always a "but" in these cases!:)
1. If this worked, casinos would have closed down! True, but this doesn't say were the problem is, i.e. what is the factor or factors that make this method non-succcessful.
2. There are betting limitations, different in ach casino (which you can see near the roulette table). The most important of them in our case is maximum bet. The lower this limit is the more chances are that you may loose a lot of chips because you couldn't double the bet anymore! (BTW, it is assumed that there's limit of the amount of money you can bet.)
3. Each roulette round lasts, what?, 5 minutes (from that start of bettings to the stop of the ball on the roulette wheel. Earning one chip each five mminutes is quite tiresome ... 6 chips in about one hour!
4. You will be eventually thrown out of the casino!
5. At a certain point of an unlucky streak, the betting sum will be too large compared to the expected gain (one chip).
6. Finally, the probability of going bankrupt, as small as this may be, compared to the small amount of money you can earn, is quite discouraging.
But you can always do that for fun ...
What is better than eternal bliss? Nothing. But a slice of bread is better than nothing. So a slice of bread is better than eternal bliss.
Your mission is not to accept the mission. Do you accept?
(Variation: Your order is not to execute the order. Would you execute it?)
Answer truthfully ("Yes" or "No"): Will the next word you say be "No"?
If the temperature this morning is 0 degrees and the Weather Channel says, "It will be twice as cold tomorrow", what will the temperature be?
(This is very interesting ... Even with -2 degrees ... -4 degrees does not feel twice as cold, it's only slightly colder.)
Everyday life statements:
• Nobody goes to that restaurant; it's too crowded.
• Don't go near the water 'til you have learned how to swim.
• The man who wrote such a stupid sentence can not write at all.
I will be glad to receive your comments and suggestions at email@example.com


For Entry Level Certificate (ELC) Mathematics at Edith Kay School, we follow the AQA specification. This qualification is helpful for our students to prepare for GCSE. The component-based structure of the qualification provides our students with the opportunity to work in short programmes. There are eight components:
This qualification is linear, where students submit all the eight components that form the assessment at the end of this course.
At Edith Kay School ELC lessons are interactive, challenging and fun – with all students fully engaged. There are plenty of maths resources available for our students to help picture abstract maths concepts in the real world, including:
Each student will complete a portfolio containing all of the eight components of work made up of between four and eight external assignments. Any remaining components will be made up of internally set classwork.
All components are internally assessed by the teacher and then moderated by AQA. Each component is marked out of 30, giving a total mark out of 240 for the whole portfolio. The outcomes are awarded from level 1 to level 3.


Police Practice Papers Number Tests 7 – 12 is additional preparation material for those who want to develop their number skills. It includes the following:
- Preparation techniques for improving number skills
- Worked examples on how to tackle each of the questions in the numbers test
- Six number tests with answers
As Numbers is the test paper which most candidates fail, a second volume of tests has been developed to allow for additional practice. Both numbers books contain a comprehensive introduction with worked examples and a suggested method of how to work through each question.


I am currently a PhD student under supervision of Stefan Schreieder at Leibniz Universität Hannover. Since October 2023, I receive a doctoral scholarship from the Studienstiftung des deutschen Volkes.
I am working in algebraic geometry and my research interests are mainly algebraic cycles and rationality questions.
Institute of Algebraic Geometry
Leibniz University Hannover
Rationality questions and decomposition of the diagonal (pdf)
Advisors: Stefan Schreieder and Nebojsa Pavic
Schnitttheorie und enumerative Geometrie [in German] (pdf)
Advisor: Stefan Schreieder


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Counting in 2s, 3s, 5s and 10s Task Cards
get link This set of counting task cards is fully aligned to the Year 2 National Curriculum 2014 Maths objective:
follow link Pupils should be taught to:
-count in steps of 2, 3, and 5 from 0, and in tens from any number, forward and backward
Included are 48 task cards, three to a page, complete with answer grids. There are 12 cards for each counting step; 2s, 3s, 5s and 10s. Pupils will count on in multiples of the particular number, filling in the missing numbers. When counting in tens, pupils will count on and back in tens from any 2-digit number.
Cards 1-12 counting on across multiples of 2
Cards 13-24 counting on across multiples of 3
Cards 25-36 counting on across multiples of 5
Cards 37-48 skip counting on and back in jumps of 10
Simply print, laminate and cut up the task cards. Pupils may write their answers using a whiteboard pen directly onto laminated cards or they can record their answers on the response sheet.


Milford High School Graduation Requirements:
24 credits: 4 English, 4 Math (including Algebra II), 3 Science (including Biology), 3 Social Studies (including US History), 2 Foreign Language, 1 PE, 0.5 Health, 3.5 Electives, 3.0 in a Pathway.
You can check your transcript to see what classes you have completed on Home Access - go to Grades and then Transcripts.
Milford High School GPA Calculation:
For each final grade on your transcript, you must convert it to quality points and then find the average with all final grades from all courses.
Weighted quality points must be used with an honors, AP, or dual enrollment course.


See how much you can save over time. Use this calculator any time you need motivation.
About Savings Calculator
A savings calculator helps you determine the amount you need to save each month. You can enter your initial deposit, the amount that you will save each month, and the interest rate that you anticipate. Then, you can enter the amount of time that you expect to have the money in your account. Then, you can see how much your savings will be worth when it has accumulated to a certain amount. Using a savings calculator will help you plan your spending based on the estimated value of your account at the end of the period you chose.
Inflation affects the value of money. When it is high, the real interest rate is lower. The nominal gains from the investment are not enough to offset the real losses. To calculate the amount of money you need to save for a specified amount of time, use a savings calculator. It’s a good idea to use this tool regularly to make sure you’re making enough money every month. Once you’ve set up your account, you can input the inflation rate to get an estimate of how much you can save each month.
Using a savings calculator can help you plan the best way to deposit your money. With these tools, you can see how much you need to save each month and the interest will accumulate. It’s important to remember that interest in a savings account builds on itself, so if you don’t want to make any monthly deposits, you can just run the numbers. A small deposit every month will move you closer to your goal.
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When you’ve set your goals, you can use a savings calculator to find the best savings account. This will help you determine the amount to deposit each month to reach your savings goal. Some savings calculators will also show you how much money you need to save in a specific time frame to meet your goals. These calculations can help you choose the best savings account for your future. If you want to save more, set up an emergency fund for an unexpected expense.
A savings calculator can also help you set your goals. This tool can help you determine how much you need to save for your future. The amount you save will depend on your personal financial situation. Once you’ve set your goal, you can use a savings calculator to figure out the amount of money you need to save. It is important to remember that you should never borrow money to pay off a debt. It’s important to pay off your debts as early as possible.


Q: What size tablecloth do you need for your table?
A: It is easy to calculate the proper size of a tablecloth:
· For a square or oblong table:
o Measure the length and width of the table.
o Decide on the length of the drop you prefer. The drop is the amount of fabric that hangs down from the top of the table. A typical drop ranges between 10 to 12”. If you want your tablecloth to go to the floor, then the drop would usually be 30” (a typical table height), but measure just in case.
o Add the amount of the drop multiplied by two to both the length and the width of the table.
o For example: If the table measures 42” wide x 84" long and you want a 12” drop, then the oblong tablecloth should measure 66” (42” width + 24” total drop) x 108” (84” length + 24” total drop).
· For a round table:
o Measure the diameter of your table.
o Decide on the length of the drop.
o Add the amount of the drop multiplied by two to the diameter of the table.
o For example: If the table measures 60” in diameter and you want it to go to the floor with a 30” drop, then the round tablecloth should measure 120” (60” diameter + 60” total drop).


The inaugural meeting of the Maths Fellowship Club will take place on Saturday 10 June 2017. The schedule for the day will be as follows:
11:00 Arrival at College for tea in the Langford Room
11:30 A talk by Professor Dominic Vella (Tutorial Fellow in Applied Mathematics) in the Oakeshott Room
12:30 Pre-lunch drinks in the Langford Room
13:00 Lunch in the Beckington Room
After lunch Tour of the Mathematical Institute (The Andrew Wiles Building)
The price of the event is £30 and places can be booked here. Please note that spaces are limited.
The Maths Fellowship Club has been established to recognise the interest that many of our alumni retain in the subject, and to encourage a meeting of minds between Fellows, students, and alumni. Our Fellowship Club events will provide an opportunity for alumni to convene in College for annual meetings, as well as to hear about our Fellows’ research, and to talk about shared interests. We are also seeking funds to support one of our Fellowship positions in Maths, a subject which is not currently endowed at Lincoln and, as such, is vulnerable should any of our current Fellows retire.