Mental calculation

# Overview
Mental calculation is the practice of doing mathematical calculations using only the human brain, with no help from any computing devices.
Practically, mental calculations are not only helpful when computing tools are not available, but they also can be helpful in situations where it is beneficial to calculate with speed. When a method is much faster than the conventional methods (as taught in school), it may be called a shortcut. Although used to aid or quicken tedious computation, many also practice or create such tricks to impress their peers with their quick calculating skills.
Almost all such methods make use of the decimal numeral system. It should be noted that the choice of radix determines what methods to use and also which calculations are easier to perform mentally. For example, multiplying or dividing by ten is very easy in decimal (just move the decimal point), whereas multiplying or dividing by sixteen is not; but the opposite happens if one uses the hexadecimal base instead of decimal.
There are many different techniques for performing mental calculations, many of which are specific to a type of problem.

# Sanity tests
A quick test to further increase confidence that the correct answer to a calculation has been found.

## Casting out nines
After applying an arithmetic operation to two operands and getting a result, you can use this procedure to improve your confidence that the result is correct.
- Sum the digits of the first operand; any 9s (or sets of digits that add to 9) can be counted as 0.
- If the resulting sum has two or more digits, sum those digits as in step one; repeat this step until the resulting sum has only one digit.
- Repeat steps one and two with the second operand. You now have two one-digit numbers, one condensed from the first operand and the other condensed from the second operand.  (These one-digit numbers are also the remainders you would end up with if you divided the original operands by 9; mathematically speaking, they're the original operands modulo 9.)
- Apply the originally specified operation to the two condensed operands, and then apply the summing-of-digits procedure to the result of the operation.
- Sum the digits of the result you originally obtained for the original calculation.
- If the result of step 4 does not equal the result of step 5, then the original answer is wrong. If the two results match, then the original answer may be right, though it isn't guaranteed to be.
- Say we've calculated that 6338  79 equals 500702
- Sum the digits of 6338: (6 + 3 = 9, so count that as 0) + 3 + 8 = 11
- Iterate as needed: 1 + 1 = 2
- Sum the digits of 79: 7 + (9 counted as 0) = 7
- Perform the original operation on the condensed operands, and sum digits: 2  7 = 14; 1 + 4 = 5
- Sum the digits of 500702: 5 + 0 + 0 + (7 + 0 + 2 = 9, which counts as 0) = 5
- 5 = 5, so there's a good chance that we were right that 6338  79 equals 500702.
You can use the same procedure with multiple operands; just repeat steps 1 and 2 for each operand.

## Estimation
When checking the mental calculation, it is useful to think of it in terms of scaling. For example, when dealing with large numbers, say 1531  19625, estimation instructs you to be aware of the number of digits expected for the final value. A useful way of checking is to estimate. 1531 is around 1500, and 19625 is around 20000, so therefore a result of around 20000  1500 (30000000) would be a good estimate for the actual answer (30045875). So if the answer has too many digits, you know you've made a mistake.

## Factors
When multiplying, a useful thing to remember is that the factors of the operands still remain. For example, to say that 14  15 was 211 would be unreasonable. Since 15 was a multiple of 5, so should the product. The correct answer is 210.

# Calculating differences: a  b

## Direct calculation
When the digits of b are all smaller than the digits of a, the calculation can be done digit by digit. For example, evaluate 872  41 simply by subtracting 1 from 2 in the units place, and 4 from 7 in the tens place: 831.

## Indirect calculation
When the above situation does not apply, the problem can sometimes be modified:
- If only one digit in b is larger than its corresponding digit in a, diminish the offending digit in b until it is equal to its corresponding digit in a. Then subtract further the amount b was diminished by from a. For example, to calculate 872  92, turn the problem into 872  72 = 800. Then subtract 20 from 800: 780.
- If more than one digit in b is larger than its corresponding digit in a, it may be easier to find how much must be added to b to get a. For example, to calculate 8192  732, we can add 8 to 732 (resulting in 740), then add 60 (to get 800), then 200 (for 1000). Next, add 192 to arrive at 1192, and, finally, add 7000 to get 8192. Our final answer is 7460.
- It might be easier to start from the left (the big numbers) first.
You may guess what is needed, and accumulate your guesses. Your guess is good as long as you haven't gone beyond the "target" number.
8192-732, mentally, you want to add 8000 but that would be too much, so we add 7000, then 700 to 1100, is 400 (so far we have 7400), and 32 to 92 can easily be recognized as 60. The result is 7460.

## Look-ahead borrow method
This method can be used to subtract numbers left to right, and if all that is required is to read the result aloud, it requires little of the user's memory even to subtract numbers of arbitrary size.
One place at a time is handled, left to right.

# Calculating products: a  b
Many of these methods work because of the distributive property.

## Multiplying by 2 or other small numbers
Where one number being multiplied is sufficiently small to be multiplied with ease by any single digit, the product can be calculated easily digit by digit from right to left. This is particularly easy for multiplication by "2" since the carry digit cannot be more than 1.
For example, to calculate 2  167: 2x7=14, so the final digit is 4, with a "1" carried and added to the 2x6=12 to give 13, so the next digit is 3 with a "1" carried and added to the 2x1=2 to give 3. Thus, the product is 334.

## Multiplying by 5
To multiply a number by 5,
1. First multiply that number by 10, then divide it by 2.
The following algorithm is a quick way to produce this result:
2. Add a zero to right side of the desired number.     (A.) 
3. Next, starting from the leftmost numeral, divide by 2 (B.)and append each result in the respective order to form a new number;(fraction answers should be rounded down to the nearest whole number).
The resulting number is 0330. (This is not the final answer, but a first approximation which will be adjusted in the following step:)
EXAMPLE: 176 (IN FIRST, SECOND THIRD PLACES):

## Multiplying by 9
Since 9 = 10  1, to multiply by 9, multiply the number by 10 and then subtract the original number from this result. For example, 9  27 = 270  27 = 243.

### Using hands: 1-10 multiplied by 9
Hold hands in front of you, palms facing you. Assign the left thumb to be 1, the left index to be 2, and so on all the way to right thumb is ten. Each | symbolizes a raised finger and a - represents a bent finger.
Bend the finger which represents the number to be multiplied by nine down.
Ex: 6  9 would be
The right little finger is down. Take the number of fingers still raised to the left of the bent finger and append it to the number of fingers to the right.
Ex: There are five fingers left of the right little finger and four to the right of the right little finger. So 6  9 = 54.

## Multiplying by 10 (and powers of ten)
To multiply an integer by 10, simply add an extra 0 to the end of the number.
To multiply a non-integer by 10, move the decimal point to the right one digit.
In general for base ten, to multiply by 10n (where n is an integer), move the decimal point n digits to the right. If n is negative, move the decimal |n| digits to the left.

## Multiplying by 11
For single digit numbers simply duplicate the number into the tens digit, for example: 1  11 = 11, 2  11 = 22, up to 9  11 = 99.
The product for any larger non-zero integer can be found by a series of additions to each of its digits from right to left, two at a time.
First take the ones digit and copy that to the temporary result. Next, starting with the ones digit of the multiplier, add each digit to the digit to its left. Each sum is then added to the left of the result, in front of all others. If a number sums to 10 or higher take the tens digit, which will always be 1, and carry it over to the next addition. Finally copy the multipliers left-most (highest valued) digit to the front of the result, adding in the carried 1 if necessary, to get the final product.
In the case of a negative 11, multiplier, or both apply the sign to the final product as per normal multiplication of the two numbers.
A step-by-step example of 759  11:
- The ones digit of the multiplier, 9, is copied to the temporary result.
result: 9
- result: 9
- Add 5 + 9 = 14 so 4 is placed on the left side of the result and carry the 1.
result: 49
- result: 49
- Similarly add 7 + 5 = 12, then add the carried 1 to get 13. Place 3 to the result and carry the 1.
result: 349
- result: 349
- Add the carried 1 to the highest valued digit in the multiplier, 7+1=8, and copy to the result to finish.
Final product of 759  11: 8349
- Final product of 759  11: 8349
Further examples:
Note the handling of 9+1 as the highest valued digit.
- Note the handling of 9+1 as the highest valued digit.
Another method is to simply multiply the number by 10, and add the original number to the result.
For example:

## Multiplying two 2 digit numbers between 11 and 19
To easily multiply 2 digit numbers together between 11 and 19 a simple algorithm is as follows:

## Multiplying Any 2 digit Numbers Together
To easily multiply any 2 digit numbers together a simple algorithm is as follows:
For example
Note that this is the same thing as the conventional sum of partial products, just restated with brevity. To minimize the number of elements being retained in one's memory, it may be convenient to perform the sum of the "cross" multiplication product first, and then add the other two elements:
i.e., in this example
to which is it is easy to add 21: 281 and then 800: 1081

## Using hands: 6-10 multiplied by another number 6-10
This technique allows a number from 6 to 10 to be multiplied by another number from 6 to 10. This is only a test.
Assign 6 to the little finger, 7 to the ring finger, 8 to the middle finger, 9 to the index finger, and 10 to the thumb. Touch the to desired numbers together. The point of contact and below is considered the "below" section and everything above the two fingers that are touching are part of the "above" section. For example, 6  9 would look like this:
Here are two examples:
above:
below:
- 5 fingers below make 5 tens
- 4 fingers above to the right
- 1 finger above to the left
the result: 9  6 = 50 + 4  1 = 54
above:
below:
- 4 fingers below make 4 tens
- 2 fingers above to the right
- 4 fingers above to the left
result: 6  8 = 40 + 2  4 = 48
How it works: each finger represents a number (between 6 and 10). Join the fingers representing the 
numbers you wish to multiply (x and y). The fingers below give the number of tens, that is (x  5) + (y  5). The digits to the upper left give (10  x) and those to the upper right give (10  y), leading to   10 + (10  x)  (10  y) = x  y.

## Using square numbers
The products of small numbers may be calculated by using the squares of integers; for example, to calculate 13  17, you can remark 15 is the mean of the two factors, and think of it as (15  2)  (15 + 2), i.e. 152  22. Knowing that 152 is 225 and 22 is 4, simple subtraction shows that 225  4 = 221, which is the desired product.
This method requires knowing by heart a certain number of squares:

## Squaring numbers
It may be useful to be aware that the difference between two successive square numbers is the sum of their respective square roots. Hence if you know that 12*12=144 and wish to know 13*13, calculate 144 + 12 + 13 = 169.
This is because (x+1)2 - x2 = x2 + 2 x + 1 - x2 = x + (x+1)
x2 = (x-1)2 + (2x - 1)

### Squaring numbers near 50
Suppose we need to square a number x near 50. This number may be expressed as x=50-n, and hence the answer x2 is (50n)2, which is 502  100n + n2. We know that 502 is 2500. So we subtract 100n from 2500, and then add n2. Example, say we want to square 48, which is 50  2. We subtract 200 from 2500 and add 4, and get x2 = 2304. For numbers larger than 50 (x=50+n), add n a hundred times instead of subtracting it.

### Squaring a number ending in 5
- Take the digit(s) that precede the five: abc5, where a, b,  and c are digits
Multiply this number by itself plus one: abc(abc + 1)
Take above result and attach 25 to the end
Example: 85  85
So, 852 = 7,225
Example: 1252
So, 1252 = 15,625
Mathematical explanation
(10x + 5)2 = 100x(x + 1) + 25
(10x + 5)(10x + 5) = 100(x2 + x) + 25
100x2 + 100x + 25 = 100x2 + 100x + 25
- Take the digit(s) that precede the five: abc5, where a, b,  and c are digits
- Multiply this number by itself plus one: abc(abc + 1)
- Take above result and attach 25 to the end
- Example: 85  85
So, 852 = 7,225
- So, 852 = 7,225
- Example: 1252
So, 1252 = 15,625
- So, 1252 = 15,625
- Mathematical explanation
(10x + 5)2 = 100x(x + 1) + 25
(10x + 5)(10x + 5) = 100(x2 + x) + 25
100x2 + 100x + 25 = 100x2 + 100x + 25
- (10x + 5)2 = 100x(x + 1) + 25
- (10x + 5)(10x + 5) = 100(x2 + x) + 25
- 100x2 + 100x + 25 = 100x2 + 100x + 25

### Squaring an integer from 26 to 75
This method requires the memorization of squares from 1 to 25.
The square of n (most easily calculated when n is between 26 and 75 inclusive) is
(50  n)2 + 100(n  25)
In other words, the square of a number is the square of its difference from fifty added to one hundred times the difference of the number and twenty five. For example, to square 62, we have:

### Squaring an integer from 76 to 99
This method requires the memorization of squares from 1 to 25.
The square of n (most easily calculated when n is between 76 and 99 inclusive) is
(100  n)2 + 100(100  2(100  n))
In other words, the square of a number is the square of its difference from one hundred added to the product of one hundred and the difference of one hundred and the product of two and the difference of one hundred and the number. For example, to square 93, we have:
Another way to look at it would be like this:
Another example:

### Squaring any 2-digit integers
This method requires memorization of the squared numerals 1 to 9.
The square of mn, mn being a two-digit integer, can be calculated as
10  m(mn + n) + n2
Meaning the square of mn can be found by adding n to mn, multiplied by m, adding 0 to the end and finally adding the square of n.
For example, we have 232:
So 232 = 529.

# Finding roots

## Approximating square roots
An easy way to approximate the square root of a number is to use the following equation:
\mathrm{root \simeq known\,\,square\,\,root - \frac{known\,\,square - unknown\,\,square}{2 - known\,\,square\,\,root}}\,\!
The closer the known square is to the unknown, the more accurate the approximation. For instance, to estimate the square root of 15, we could start with the knowledge that the nearest perfect square is 16 (42).
\begin{align}
\mathrm{root} & \simeq 4 - \frac{16 - 15}{2 - 4} \\
& \simeq 4 - 0.125 \\
& \simeq 3.875 \\
\end{align}\,\!
So we've estimated the square root of 15 to be 3.875. The actual square root of 15 is 3.872983...
Derivation
Say we want to find the square root of a number we'll call 'x'. By definition
\mathrm{root}^2 = x\,\!
We then redefine the root
\mathrm{root} = a - b\,\!
where 'a' is a known root (4 from the above example) and 'b' is the difference between the known root and the answer we seek.
(a-b)^2 = x\,\!
Expanding yields
a^2 - 2ab + b^2 = x\,\!
And here's the trick. If 'a' is close to your target, 'b' will be a small enough number to render the +\,b^2\,\! element of the equation negligible. So we drop +\,b^2\,\! out and rearrange the equation to
b \simeq \frac{a^2 - x}{2a}\,\!
and therefore
\mathrm{root} \simeq a - \frac{a^2 - x}{2a}\,\!

## Extracting roots of perfect powers
This is a surprisingly easy task for many higher powers, but not very useful except for impressing friends (practical uses of finding roots rarely use perfect powers). The task is not as hard as it sounds mainly because the basic method is to find the last digit using the last digit of the given power and then finding the other digits by using the magnitude of the given power. Such feats may seem obscure but are nevertheless recorded and practiced. See 13th root.

### Extracting cube roots
An easy task for the beginner is extracting cube roots from the cubes of 2 digit numbers. For example, given 74088, determine what two digit number, when multiplied by itself once and then multiplied by the number again, yields 74088. One who knows the method will quickly know the answer is 42, as 423 = 74088.
Before learning the procedure, it is required that the performer memorize the cubes of the numbers 1-10:
A neat trick here is that there is a pattern see if you can see it. Remember that the pattern is adding and subtracting. Starting from zero:
- 13 = 1 up 1
- 23 = 8 down 3
- 33 = 27 down 1
- 43 = 64 down 3
- 53 = 125 up 1
- 63 = 216 up 1
- 73 = 343 down 3
- 83 = 512 down 1
- 93 = 729 down 3
- 103 = 1000 up 1
There are two steps to extracting the cube root from the cube of a two digit number. Say you are asked to extract the cube root of 29791. Begin by determining the one's place (units) of the two digit number. You know it must be one, since the cube ends in 1, as seen above.
- If perfect cube ends in 0, the cube root of it must end in 0.
- If perfect cube ends in 1, the cube root of it must end in 1.
- If perfect cube ends in 2, the cube root of it must end in 8.
- If perfect cube ends in 3, the cube root of it must end in 7.
- If perfect cube ends in 4, the cube root of it must end in 4.
- If perfect cube ends in 5, the cube root of it must end in 5.
- If perfect cube ends in 6, the cube root of it must end in 6.
- If perfect cube ends in 7, the cube root of it must end in 3.
- If perfect cube ends in 8, the cube root of it must end in 2.
- If perfect cube ends in 9, the cube root of it must end in 9.
Note that every digit corresponds to itself except for 2, 3, 7 and 8, which are just subtracted from ten to obtain the corresponding digit.
The second step is to determine the first digit of the two digit cube root by looking at the magnitude of the given cube. To do this, remove the last three digits of the given cube (29791 -> 29) and find the greatest cube it is greater than (this is where knowing the cubes of numbers 1-10 is needed). Here, 29 is greater than 1 cubed, greater than 2 cubed, greater than 3 cubed, but not greater than 4 cubed. The greatest cube it is greater than is 3, so the first digit of the two digit cube must be 3.
Therefore, the cube root of 29791 is 31.
Another example:
- Find the cube root of 456533.
- The cube root ends in 7.
- After the last three digits are taken away, 456 remains.
- 456 is greater than all the cubes up to 7 cubed.
- The first digit of the cube root is 7.
- The cube root of 456533 is 77.

### Finding Fifth Roots
Nth root

# Approximating common logs (log base 10)
To approximate a common log (to at least one decimal point accuracy), a few log rules, and the memorization of a few logs is required. One must know:
- log(ab) = log(a)+log(b)
- log(a/b) = log(a)-log(b)
- log(0) does not exist
- log(1) = 0
- log(2) ~ .30
- log(3) ~ .48
- log(7) ~ .85
From this information, one can find the log of any number 1-9.
- log(1) = 0
- log(2) ~ .30
- log(3) ~ .48
- log(4) = log(2*2) = log(2)+log(2) ~ .60
- log(5) = log(10/2) = log(10)-log(2) ~ .70
- log(6) = log(2*3) = log(2)+log(3) ~ .78
- log(7) ~ .85
- log(8) = log(2*2*2) = log(2)+log(2)+log(2) ~ .90
- log(9) = log(3*3) = log(3)+log(3) ~ .96
- log(10) = 1+log(1) = 1
The first step in approximating the common log is to put the number given in scientific notation. For example, the number 45 in scientific notation is 4.5 - 10^1, but we will call it a - 10^b. Next, find the log of a, which is between 1 and 10. Start by finding the log of 4, which is .60, and then the log of 5, which is .70 because 4.5 is between these two. Next, and skill at this comes with practice, place a 5 on a logarithmic scale between .6 and .7, somewhere around .653 (NOTE: the actual value of the extra places will always be greater than if it were placed on a regular scale. i.e., you would expect it to go at .650 because it is halfway, but instead it will be a little larger, in this case .653)  Once you have obtained the log of a, simply add b to it to get the approximation of the common log. In this case, a+b = .653+1 = 1.653. The actual value of log(45) = 1.65321.
The same process applies for numbers between 0 and 1. For example, .045 would be written as 4.5*10^-2. The only difference is that b is now negative, so when adding you are really subtracting. This would yield the result .653-2, or -1.347.

# Other systems
There are many other methods of calculation in mental mathematics. The list below shows a few other methods of calculating, though they may not be entirely mental.
- Vedic mathematics
- Trachtenberg system
- Abacus system
- Chisanbop

# Mental Calculation World Cup
The first World Mental Calculation Championships (Mental Calculation World Cup) took place 2004. They are repeated every second year. The event of 2006 took place on November 4, 2006 in Giessen, Germany. It consists of four different tasks: addition of ten ten-digit numbers, multiplication of two eight-digit numbers, calculation of square roots and calculation of weekdays for given dates, plus two surprise tasks. It was won by Robert Fountain from England.
The next Championship is scheduled for 2008.