How to Do Cube Roots of 9 Digit Numbers in Your Head How to Do Cube Roots of 9 Digit Numbers in Your Head 19 September, 2005 cube root of a 9 digit number To get the first digit, calculate the cube root of the first 3 digits. To get the last digit, calculate the cube root mod 10. To get the middle digit, calculate the cube root mod 11. tags: Update (23 April 2006): try out the new Practice Page What is a Cube Root? (Just in case you don't already know.) A number cubed is that number times itself times itself. For example, 4 cubed, normally written 43, is 4 × 4 × 4 = 64. The cube root goes in the opposite direction, so the cube root of 64 is 4. To give a larger example, 657 cubed = 6573 = 657 × 657 × 657 = 283593393, so the cube root of 283593393 is 657.

Doing it in your head Now I'll have to admit that I used a calculator to calculate that 6573 = 283593393 (actually I'm lying – I did it in Python , but that's beside the point). But I am claiming that you can learn to calculate the cube root of numbers like 283593393, by mental arithmetic alone. First Cheat When I say "the cube root of numbers like 283593393", there is of course something special about 283593393, which is that it is a cube of a whole number, so it's cube root is a whole number. Although there are 999999999 numbers up to 9 digits long, only 999 of these numbers have cube roots that are whole numbers. But if you are giving a display of your mental arithmetic powers to an audience, this is no problem. Give someone in your audience a calculator. Tell them to think of a 3 digit number, multiply it by itself twice and tell you the answer (which will be 7 to 9 digits long).

You will then tell them what the original number was. Don't ask them to supply a random 9 digit number, because a random 9 digit number probably won't be a cube of a whole number, and you will be in trouble, and the following strategies won't work. Cube Roots up to 3 Digits If you had to calculate the cube root of a 3 digit number, it would be quite easy. Here is a table of cubes from 1 to 10: n 10 1000 To do cube roots of 9 digit numbers, you are going to have to be familiar with these first 10 cubes. So you might as well memorise them. And if you have memorised them, then there isn't really any trick to calculating the cube roots. Cube Roots up to 6 Digits What is the cube root of 300763? The cube root of a 6 digit number has to be less than 100, so you could consider memorising the first 100 cubes. However that is a bit much, and it is also unnecessary, so I won't suggest that you memorise that many cubes.

Let's look at the first 3 digits, which are "300". Now 300 is greater than 216 = 63 and less than 343 = 73. Which means that 300763 is greater than 216000 = 603 and less than 343000 = 703. So we know that our answer is somewhere between 60 and 70. To get the last digit easily, we need to use modular arithmetic, and to make it really easy, we will do arithmetic modulo 10. All this means is doing addition, subtraction and multiplication, keeping only the last digit of the answer. A modulo 10 version of the previous table is as follows: n 0 0 A curious pattern emerges: if we cube a number modulo 10 twice, we get back to where we started. So 83 = 512 = 2 (mod 10, where "mod" is an abbreviation for "modulo"), and 23 = 8 = 8 (mod 10).

Applying this pattern to our problem cube 300763, we deduce that the last digit must be 33 = 27 = 7 (mod 10). So the answer must be a number in between 60 and 70 with last digit 7, which can only be 67. Cube Roots up to 9 Digits Using the same techniques, we can easily calculate the first and last digits of the cube root of a 9 digit number (assuming, as already stated, that it is known to be the cube of a whole number). So, for example, looking at 580093704, we compare 580 to 512 and 729, to deduce that the cube root is between 800 and 900. The last digit is 4, so the last digit of the cube root must be 43 = 64 = 4 (mod 10). All that is left to find is the middle digit. Unfortunately we have extracted all the information we can out of size comparisons and mod 10 arithmetic, and we need to think of something else.

One "something else" is a different modulus. However this means doing long divisions in your head to get remainders, which is possibly a bit much work. We need a modulus for which there is an easy shortcut. And it has to be bigger than 10, if it is to uniquely choose the one correct answer from 10 possibilities. There are two major candidates, which are 9 and 11. Modulo 9 is easy to calculate, because you calculate the remainder of a number divided by 9 by just adding the digits. For example 580093704 = 5+8+0+0+9+3+7+0+4 = 36 mod 9 = 3+6 (mod 9) = 9 (mod 9) = 0. One problem with 9 is that it isn't bigger than 10. However, this might not be completely fatal. There is a short cut for modulo 11, but it is a little harder. Starting from the right end, you alternatively add and subtract digits.

So 580093704 = 4 - 0 + 7 - 3 + 9 - 0 + 0 - 8 + 5 = 14 = 4 - 1 = 3 (modulo 11). (In this case we luckily ended up with a positive number less than 11, but you might have to adjust the answer to a number in this range by adding or subtracting 11.) To make use of these moduli, we need to calculate the tables of cubes mod 9 and 11: n 1 10 Unfortunately there is a problem with mod 9, which is that there are only 3 possible cubes mod 9, which means that going the other way there are still going to be 3 or 4 choices. Mod 11 works better, and we can see that each value from 0 to 10 occurs only once in the list of cubes modulo 11 (from 0 to 10), which means that cube roots can be calculated modulo 11.

Returning back to the worked example, 580093704 = 3 modulo 11, so (looking at the table, which, by the way, you will have to memorise) its cube root = 9 modulo 11. We've already determined that the first digit is 8 and the last digit is 4. If we think of the cube root as 8x4, for some x, then we have the equation 4 - x + 8 = 9 modulo 11, i.e. 12 - x = 9 modulo 11, so x = 3 modulo 11. Which gives a final (and correct) answer of 834. Conclusion So there you have it, how to calculate cube roots of 9 digit numbers in your head. There is a certain amount of memorisation and fiddly addition and subtraction to do, and you probably need to be confident with these before you try it out as a party trick.

I suggest that it will also be much easier if you make sure that you can see the number you are calculating the cube root of, so that you are not saddled with the additional mental load of having to memorise that number. If you liked this page ... If you liked this page, you might also like my PrimeShooter Vedic Math - Cube of 2 Digit Numbers ~ VedantaTree ~ The Tree of Knowledge Vedic Math Vedic Math - Cube of 2 Digit Numbers Posted by Pooja Gupta on Sunday, February 3, 2013 Note: Vedic Math Blog has been moved to . Please bookmark the new address for new and existing blogs. Cube of numbers plays an important role in mathematics calculations, like in finding cube root of the numbers. So it is useful if we can do the cube of numbers quickly. We are trying for the same here using 'Anurupya Sutra' of Vedic Math. Let us learn how we can do it. Cubes of the single digits i.e.

from (1 to 9) are given below: 13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125, 63 = 216, 73 = 343, 83 = 512, 93 = 729, 103 = 1000 If we observe closely, the last digit of every cubic number is unique i.e. numbers from (1 to 9) does not repeat. This observation will be very helpful while calculating cube roots of any 2-digit numbers. Let us first see the Algebraic Expression for Cube root: (a + b)3 = a3 + 3a2b + 3ab2 + b3 Above expression for cube root of (a + b) contain 4 terms in total.

1st term is a3 2nd term is a2b = a3 x (b/a) = 1st term x (b/a) 3rd term is ab2 = a2b x (b/a) = 2nd term x (b/a) 4th term is b3 = ab2 x (b/a) = 3rd term x (b/a) Here (b/a) is the common ratio Also, as the whole, 2nd term is 3a2b = a2b + 2a2b {split as sum of two terms} and, 3rd term is 3ab2 = ab2 + 2ab2 {split as sum of two terms} So to find the cube, we have to compute a3 and b/a. In Vedic Math, same formula can be used in a different way to find the cube of 2-digit numbers i.e. ab. Apply formula on 'ab' like (a+b)3 as stated above, and add the results of different rows in vertical columns. You will be able to do the cube of any two digit numbers quickly.

We shall use 'Anurupya Sutra' of Vedic Math for this cube calculation, which states: "If you start with the cube of first digit and take the next three numbers (in the top row) in a Geometrical Proportion (in the ratio of original digits themselves), you will find that the fourth figure on the right hand will be just the cube of second digit". Following is the step by step description of finding the cube of 2-digit number: Step 1: In the first row, start with a^3 as 1st term and multiplying 1st term by (b/a) to get 2nd term. Step 2: Repeat the multiplication till 4th term. Step 3: In the second row, double the two middle terms (i.e. 2nd term and 3rd term) and write just below 2nd term and 3rd term. Step 4: Add them vertically in columns. Carry forward the 10th place digit to next column. The example given below will describe this method well.

Example: 113 Here a = 1 , b = 1 , a3 = 1 , b/a = 1/1 = 1 (Here common ratio is equal to 1) Now see the formation of the table: First Row 1 1 1 1 Second Row 2 2 Add 1 3 3 1 113 = 1331 Example: 133 Here a = 1 , b = 3 , a3 = 1 , b/a = 3/1 = 3 (Here common ratio is greater than 1) Now see the formation of the table: First Row 1 3 9 27 (Note: 4th term is just b3 as shown above in algebraic Second Row 6 18 expression) ------------------------- Add 1 9 27 27 ------------------------- = 1 9 7 7 (Apply carry over rule) 2 2 cube root cube root (algorithm) Definition: This describes a "long hand" or manual method of calculating or extracting cube roots. Calculation of a cube root by hand is similar to long-hand division or manual square root .

Suppose you need to find the cube root of 55,742,968. Set up a "division" with the number under the radical. Mark off triples of digits, starting from the decimal point and working left. (The decimal point is a period (.), and commas (,) mark triples of digits.) ____________ \/ 55,742,968. Look at the leftmost digit(s) (55 in this case). What is the largest number whose cube is less than or equal to it? It is 3, whose cube is 27. Write 3 above, write the cube below and subtract. __3_________ Now bring down the next three digits (742). __3_________ ---- 28742 Coming up with the next "divisor" is more involved than for square roots . First bring down 3 times the square of the number on top (3 × 3²=27) leaving room for two more digits (27_ _).

__3_________ ---- 27_ _) 28742 What is the largest number that we can put in the next position and multiply times the divisor and still be less than or equal to what we have? (Algebraically, what is d such that d × 2700 ≤ 28742?) 10 might work (since 10 × 2700 = 27000), but we can only use a single digit, so we'll try 9. __3___9_____ ---- 27_ _) 28742 The second step in making the divisor is adding 3 times the previous number on top (3) times the last digit (9) times 10 (3 × 3 × 9 = 81 × 10 = 810) and the square of the last digit (9² = 81). 2700 Our new divisor is 3591. __3___9_____ ---- 3591) 28742 Multiply by the last digit (9 × 3591 = 32319) and subtract.

But that is too big! So we'll try 8 as the next digit instead. __3___8_____ ---- 27_ _) 28742 We repeat the second step of adding 3 times the previous number on top (3) times the last digit (8) times 10 (3 × 3 × 8 = 72 × 10 = 720) and the square of the last digit (8² = 64). 2700 Our new divisor is 3484. __3___8_____ Now multiply by the last digit (8 × 3484 = 27872) and subtract. __3___8_____ ------ 870 We are ready to start over on the next digit. Bring down the next three digits. The divisor starts as 3 times the square of the number on top (3 × 38²=4332) leaving room for two more digits (4332_ _). __3___8_____ It looks like 2 is the next digit.

__3___8___2_ ------ 4332_ _) 870968 Add 3 times the previous number on top (38) times the last digit (2) times 10 (3 × 38 × 2 = 228 × 10 = 2280) and the square of the last digit (2² = 4). 433200 Our new divisor is 435484. __3___8___2_ Now multiply by the last digit (2 × 435484 = 870968) and subtract. __3___8___2_ ------- 0 So the cube root of 55742968 is 382. You can continue to get as many decimal places as you need: just bring down more triples of zeros. Why does this work? Consider (10A + B)³ = 1000A³ + 3 × 100A²B + 3 × 10AB² + B³ and think about finding the volume of a cube.

The volume of the three thin plates is 3 × 100A²B. The volume of the three skinny sticks is 3 × 10AB². The tiny cube is B³. If we know A and the volume of the cube, S, what B should we choose? We previously subtracted A³ from S. To scale to 1000A³, we bring down three more digits (a factor of 1000) of the length of S. We write down 3 times A squared (3A²), but shifted two places (100 × 3A² or 3 × 100A²). We estimate B. We add 30 times A times B (30 × AB or 3 × 10AB) and B squared. Multiplying that by B gives us 3 × 100A²B + 3 × 10AB² + B³. When we subtract that from the remainder (remember we already subtracted A³), we have subtracted exactly (10A + B)³. That is, we have improved our knowledge of the cube root by one digit, B.

We take whatever remains, scale again by 1000, by bringing down three more digits, and repeat the process. See also square root . Note: In computers and hand-held calculators, square root, sine, cosine, and other transcendental functions are calculated with sophisticated functions, such as Taylor series, CORDIC , or Newton-Raphson method , sometimes called Newton's method. This lesson explains, for instance, possible difficulties in convergence . Cubes and Cube Roots Cubes and Cube Roots To understand cube roots, first we must understand cubes ... How to Cube A Number To cube a number, just use it in a multiplication 3 times ... Example: What is 3 Cubed? 3 Cubed Note: we write down "3 Cubed" as 33 (the little 3 means the number appears three times in multiplying) Some More Cubes A cube root goes the other direction: 3 cubed is 27, so the cube root of 27 is 3 3 The cube root of a number is ... ... a special value that when cubed gives the original number. The cube root of 27 is ... ... 3, because when 3 is cubed you get 27.

Note: When you see "root" think "I know the tree, but what is the root that produced it?" In this case the tree is "27", and the cube root is "3". Here are some more cubes and cube roots: 4 216 Example: What is the Cube root of 125? Well, we just happen to know that 125 = 5 × 5 × 5 (if you use 5 three times in a multiplication you will get 125) ... ... so the answer is 5 The Cube Root Symbol This is the special symbol that means "cube root", it is the "radical" symbol (used for square roots) with a little three to mean cube root. You can use it like this: (we say "the cube root of 27 equals 3") You Can Also Cube Negative Numbers Have a look at this: When we cube 5 we get 125: When we cube −5 we get −125: So the cube root of −125 is −5 Perfect Cubes The Perfect Cubes are the cubes of the whole numbers : 3375 ...

It is easy to work out the cube root of a perfect cube, but it is really hard to work out other cube roots. Example: what is the cube root of 30? Well, 3 × 3 × 3 = 27 and 4 × 4 × 4 = 64, so we can guess the answer is between 3 and 4. Let's try 3.5: 3.5 × 3.5 × 3.5 = 42.875 Let's try 3.2: 3.2 × 3.2 × 3.2 = 32.768 Let's try 3.1: 3.1 × 3.1 × 3.1 = 29.791 We are getting closer, but very slowly ... at this point, I get out my calculator and it says: 3.1072325059538588668776624275224... ... but the digits just go on and on, without any pattern. So even the calculator's answer is only an approximation !

(Further reading: these kind of numbers are called surds which are a special type of irrational number ) Cube Numbers To the main page "Mathematische Basteleien" What is a cube number? A cube number (or a cube) is a number you can write as a product of three equal factors of natural numbers. Formula: k=a*a*a=a³ (k and a stand for integers.) On the other hand a cube number results by multiplying an integer by itself three times. Formula: a*a*a=a³=k (a and k stand for integers.) The same factor is called the base. After this a negative number like (-2)³= -8 or a fraction number like (2/3)³=8/27 are suspended. If it is appropriate, the number 0 is also a cubic number. These are the first 100 cube numbers. You can illustrate the name cube number by the following drawings. If a is the side of a cube, the volume is a³. ...1... 1 Computation with the calculator In former times you found cube numbers by ready-reckoners, today there are calculators.

You act with the TI30 like this: After entering the number you press the keys yx , (1/3) and finally = . Example: [2299968] -- [yx] -- [(1/3)] -- [=] . You get 22999681/3 =132. It is even simpler to give in 2299968^(1/3) into the searching field of Google or Bing. The cube root appears after pressing the enter key. 2 Determination by nested intervals 22999681/3 The number must be between 100 and 200 (100³=1000000 and 200³=4000000). It must be between 130 and 140 (130³=2197000 and 140³=2744000). It must be near 130 and 8 is on the ones place. Then 132 comes into consideration.

Result: 22999681/3=132 22999681/3 You write the number in factors and develop the cube number this way. 2299968 = 8*287496 = 8*27*10648 = 8*27*8*1331=8*27*8*11*121=8*8*27*11³=(2*2*3*11)³=132. Result: 22999681/3 = 132. Cube Roots of Negative Numbers? top There is (-2)³=-8. You could think that then (-8)1/3=-2 is. In former times this notation was common in schools, too. Today you demand that a1/3 is only defined for a>0 or a=0. The term (-8)1/3 ist not allowed. Thus you avoid conflicts as the following calculations show.

There is (-8)1/3=(-8)2/6=[(-8)2]1/6=(64)1/6=(26 )1/6=2. That means that (-8)1/3 would be ambiguous. You would get two results depending on the way of calculation. This is possible. The equation (-2)³=-8 is changed to -(81/3)=-2. Generally x³=a leads to x=-(-a)1/3 for a<0. Sequences and Series top Series of the cube numbers There is a series to each sequence and this is the sequence of the partial sums. For cube numbers this is the sequence sn= 1³+2³+3³+...+n³. 1³+2³+3³+...+n³=[n(n+1)/2]² is given. Proof by mathematical induction >The statement holds for n=1: s1=[1*(1+1)/2]²=1. >Assume that the formula holds for all natural numbers n: sn=[n(n+1)/2]².

>You must show that then the formula also holds for n+1: sn+1 = [(n+1)(n+2)/2]². Calculation: Convert the product [(n+1)(n+2)]² into a sum at first. [(n+1)(n+2)]²=(n²+3n+2)²=n4+(3n+2)²+2n²(3n+2)=n4+(9n²+12n+4)+(6n³+4n²)=n4+6n³+12n²+12n+4 There is sn+1 = sn+(n+1)³ = [n(n+1)/2]²+(n+1)³ = [n²(n+1)²+4(n+1)³]/4 = [n²(n²+2n+1)+(4n³+12n²+12n+4)]/4 = (n4+6n³+13n²+12n+4)/4 = [(n+1)(n+2)]²/4 = [(n+1)(n+2)/2]², qed..

The formula 1³+2³+3³+...+n³=[n(n+1)/2]² contains the series of the natural numbers, because the sum can be written as 1+2+3+ ...+n=n(n+1)/2. Thus there is the formula 1³+2³+3³+...+n³ = (1+2+3+...+n)², which is illustrated on the right. Interpretation as an arithmetic series of third order The sequence of the cube numbers is also an arithmetic series of third order. A figure follows to explain this. In every new row you find the difference of two numbers a row higher. The feature is that you reach a constant difference 6 after three steps. There are formulas for these sequences, so that you can prove the formula 1³+2³+3³+...+n³=[n(n+1)/2]² in another way. You can read this on Steffen Weber's (URL below) German web site.

Series of odd numbers There is 1³=1, 2³=3+5, 3³=7+9+11, 4³=13+15+17+19, ... and 1³+2³+3³+4³+... = 1+3+5+7+9+11+13+15+17+19+... Series of the reciprocal cube numbers There is another series of cube numbers, the one of the reciprocal cube numbers: 1/1³+1/2³+1/3³+...+1/n³. It is convergent. You find the limit, the Riemann's zeta-function for 3, on the Wikipedia page 1729 (number). Series of the centered hexagonal numbers The sequence of the centered hexagonal numbers is 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, ... The formula is an=3n²-3n+1.

There is 1, 1+7=8, 1+7+19=27, 1+7+19+37=64, ... The series to the sequence of the centered hexagonal numbers is the sequence of the cube numbers. Calculation for proving: The formulas 1²+2²+3³+...+n²=(1/6)n(n+1)(2n+1) and 1+2+3+...+n=(1/2)n(n+1) are used. Sequence of the centered cube numbers The sequence of the centered cube numbers is 1, 9, 35, 91, 189, 341, 559, 855, 1241,... ...1... The picture pairs make a 3-D view possible. The formula is an=2n³-3n²+3n-1. If you add two consecutive cube numbers, the centered cube numbers develop. This is given by 2n³-3n²+3n-1=n³+(n -1)³. Atomium in Brussels, a centered cube ...

It stands 102 meters (335 ft) tall. There are nine steel spheres connected by tubes. Photos from July 2011. Sequence of the perfect numbers T.L. Heath (1861-1940) proved that every even perfect number - except 6 - is the sum of 2(n-1)/2 cube numbers, for example 28=1³+3³, 496=1³+3³+5³+7³, 8128=1³+3³+5³+7³+9³+11³+13³+15³. Source (3), page 14f. Waring's Problem top The English mathematician Eduard Waring (1734-1798) maintained the following statement among others. "Every natural number is either a cube number or the sum of 2,3,4,5,6,7,8 or 9 cube numbers." (2), page 37ff. That means that 9 is a smallest number.

It can be more than 9 as the following sum of 180³ with 64 (!) cubic numbers shows. 180³ = 6³+7³+8³+...+67³+68³+69³ (1). Already 4 summands will do, 180=1³+3³+3³+5³.

The first numbers 2=1³+1³ 3=1³+1³+1³ 4=1³+1³+1³+1³ 5=1³+1³+1³+1³+1³ 6=1³+1³+1³+1³+1³+1³ 7=1³+1³+1³+1³+1³+1³+1³ 15=2³+1³+1³+1³+1³+1³+1³+1³ 23=2³+2³+1³+1³+1³+1³+1³+1³+1³ Dissection of the numbers <= 100, found by a simple computer program The smallest number of the summands is determined.

>The numbers 2, 9, 16, 28, 35, 54, 65, 72, 91 have sums with 2 cubes at least: 2, 9, 16, 28, 35, 54, 65, 72, 91 >Sums with 3 cubes: 3, 10, 17, 24, 29, 36, 43, 55, 62, 66, 73, 80, 81, 92, 99 >Sums with 4 cubes:4, 11, 18, 25, 30, 32, 37, 44, 51, 56, 63, 67, 70, 74, 82, 88, 89, 93, 100 >Sums with 5 cubes: 5, 12, 19, 26, 31, 33,

 38, 40, 45, 52, 57, 59, 68, 71, 75, 78, 83, 90, 94, 96, 97 >Sums with 6 cubes: 6, 13, 20, 27, 34, 39, 41, 46, 48, 53, 58, 60, 69, 76, 79, 84, 86, 95, 98 >Sums with 7 cubes: 7, 14, 21, 42, 47, 49, 61, 77, 85, 87 >Sums with 8 cubes: 15, 22, 50 (All numbers are 15, 22, 50, 114, 167, 175, 186, 212, 231, 238

, 303, 364, 420, 428, 454.) >Sums with 9 cubes for 23, (and 239 only) >8, 27, 64 are the cube numbers < 100.

There is an anecdote to the sums 1729=1³ + 12³ and 1729= 9³ + 10³, which you can read on the German Wikipedia site Tausendsiebenhundertneunundzwanzig (URL below). Sum of cubes The equation x²+y²=z² leads to the Pythagorean triples (more on my German page 3-4-5-Dreieck ) The more general equation xn+yn=zn has no solution for n>2. This is the famous Fermat's Last Theorem. Andrew Wiles und Richard Taylor proved the theorem not until 1993/95. I recommend the fascinating story in Simon Singh's bestseller Fermat's Last Theorem (5). If you cannot write a cube number as a sum of two cubes, you can look for sums with three or more cubes. A nice example is 3³+4³+5³=6³, particularly, because there are four consecutive numbers. There are more examples in the chapter Playings with Numbers below. Special Cube Numbers top Square numbers among the cubes There are cubes, which also are squares.

You can construct them step by step by squaring cube numbers. 23 leads to 26=64, 33 to 36=729, 4³ to 46=4 096, ... The next cube numbers and at the same time square numbers upto 1 million are 15 625, 46 656, 117 649, 262 144 und 531 441. I must not forget 1. Palindromes among the cube numbers 343=7³ 131=11³, 1030301=101³, 1003003001=1001³, ... 1367631=111³, 1030607060301=10101³, 1003006007006003001=1001001³, ...

1334996994331=11011³, 1331399339931331=110011³, ... 1033394994933301=1011010³, ... It is noticed that all bases are also palindromes. The only palindrome with a non-palindrome base is 10662526601 = 2201³ according to Martin Gardner. Source: Patrick De Geest (World!of numbers) Playing with Numbers top I show results of Lietzmann's book (1) from 1948 among other things in this chapter, extended and controlled by my computer. A cube number is the third power of its digit sum.

512=8³=(5+1+2)³ 4913=17³=(4+9+1+3)³ 5832=18³=(5+8+3+2)³ 17576=26³=(1+7+5+7+6)³ 19683=17³=(1+9+6+8+3)³ A number is the sum of the third power of its digits. 153=1³+5³+3³ 370=3³+7³+0³ 371=3³+7³+1³ 407=4³+0³+7³ The simple program on the right found this.

for x=0 to 9 for y=0 to 9 for z=0 to 9 if x*x*x+y*y*y+z*z*z=1000*x+100*y+z then print x,y,z next z 22+2=2³+2³+2³ 12*3=1³+2³+3³ 32*5=3³+2³+5³ 50*5=5³+0³+5³ 51*3=5³+1³+3³ 151+3=1³+5³+1³+3³ und 153+1 370+1=3³+7³+0³+1³ 371+1=3³+7³+1³+1³ 400+7=4³+0³+0³+7³ 401+7=4³+0³+1³+7³ und 407+1=4³+0³+1³+7³ 464+5=4³+6³+4³+5³ und 

465+4=4³+6³+4³+5³ 624+7=6³+2³+4³+7³ und 627+4=6³+2³+4³+7³ 643+7=6³+4³+3³+7³ und 647+3=6³+4³+3³+7³ 733+7=7³+3³+3³+7³ und 737+3=7³+3³+3³+7³ 773+4=7³+7³+3³+4³ und 774+3=7³+7³+3³+4³ 914+5=9³+1³+4³+5³ und 915+4=9³+1³+4³+5³ 12+32=1³+2³+3³+2³ 20+23=2³+0³+2³+3³ 21+23=2³+1³+2³+3³

 30+32=3³+0³+3³+2³ 31+32=3³+1³+3³+2³ 107*8=1³+0³+7³+8³ 180*3=1³+8³+0³+3³ 989*2=9³+8³+9³+2³ 18*30=1³+8³+3³+0³ Terms with equal digits 3³+7³=37*(3+7) 4³+8³=48*(4+8) 14³+7³=147*(14+7) 14³+8³=148*(14+8) Cube numbers are written in the digits 1 to 9.

No digit is twice or more.

125*438976=380³ 24+76=100, 13824 and 438976 25+75=100, 15625 and 421875 49+51=100, 117649 and 132651 125+875=1000, 1953125 and 669921875 251+749=1000, 15813251 and 420189749 0624+9376=10000, 242970624 and 824238309 376 0625+9375=10000, 244140625 and 823974609 375 21952 and 2+1+9+

5+2, plus 6+8+5+9 and 6859 The bases form an arithmetic progression 180³ = 6³+7³+8³+...+67³+68³+69³ 540³ = 34³+35³+ ...

+158³ 2856³ = 213³+214³+ ... +555³ 5544³ = 406³+407³+ ... +917³ 16830³ = 1134³+1135³+ ... +2133³ 3990³ = 290³+293³+ ... +935³ 29880³ = 2108³+2111³+ ...

+4292³ 408³ = 149³+256³+363³ 440³ = 230³+243³+265³+269³+282³ 1155³ = 435³+506³+577³+648³+719³+790³ 2128³ = 553³+710³+867³+1024³+1181³+1338³+1475³ 168³ = 28³+41³+54³+67³+80³+93³+106³+119³ 64085³ = 935³+5868³+10801³+15734³+20667³+25600³+30533³+35466³+40399³+45332³

 495³ = 15³+52³+89³+126³+163³+200³+237³+274³+311³+248³ Cube Numbers on the Internet top Deutsch h2g2 - How to Do Large Cube Roots in Your Head - Edited Entry To The Galaxy Earth Edition How to Do Large Cube Roots in Your Head Content from the guide to life, the universe and everything How to Do Large Cube Roots in Your Head Created 0 Conversations You will have heard of those geniuses who can remember 100,000 digits of pi , who memorise whole books or who even manage to remember Auntie Betsy's birthday.

You may not be one of them, but one thing you will certainly be able to do, to the amazement of your friends, is computing the cube roots of numbers up to one billion. Doubts? Of course you do not have to calculate them the usual way. Instead, you will learn a much simpler method in this Entry. A Refresher Course in Cubes and Cube Roots Just to remind you, a cube is when you take a number and multiply it by itself twice: that is, you take three copies of the number and multiply them together. For example, the cube of 4, or 4 cubed, is 4 × 4 × 4 which is 64. As a shorthand, we use a tiny '3' to denote a cube: 43 = 4 × 4 × 4 = 64 A cube root is the opposite of a cube: given the cube, find the number which needs to be multiplied to produce it. This is the cube root. In the example above, the cube root of 64 is 4.

64 = 4 × 4 × 4 ³√ 64 = 4 Pick a Number, Any Number Of course, most numbers have cube roots which are not nice round numbers like 4. The cube root of 100, for example, is approximately 4.641588834. The method used in this Entry will not work for such numbers. It only works for numbers which have a whole number as their cube root. The best way to ensure this, when demonstrating your skills, is to get your friend to pick a three-digit number but to keep it secret from you. Get him to cube it (preferably on a calculator). He tells you the result, then you calculate the cube root in your head, giving him his original number. Gasps all round! Cube Roots of Numbers Up to Six Digits Before tackling this, let's first take a look at numbers up to a million. Although a million already seems to be pretty large, its cube root is just 100. That means you could do the job by memorising 100 pairs of numbers.

But the following eleven will also do: 0 10 1,000 The number in the right column is the cube of the number in the left column. Conversely, the number on the left is the cube root of the number on the right. Note that each of those numbers ends with a different digit. If you cube a number with two digits ab, only the second digit, b, will affect the last digit of the cube and can therefore be deduced easily. For example, if the cube ends in 3, the cube root must end in 7. To find a, the first digit of the cube root, look at the powers of 10, 20, 30 and so on, which are those in the table above multiplied by 1,000. For example, 903 is equal to 729,000. The biggest of them that is equal or less than ab cubed is assigned to a in the table. Problem solved! Example What is the cube root of 21,952?

The cube lies between 8,000 and 27,000; so its root has to be between 20 and 30. The last digit equals 2 as in 512, so the second digit of the root must be 8. Therefore the whole root is 28. Up to Nine-Digit Cubes Moreover, with this method we can tell the first and last digit of every natural cube root regardless of its size. With this in mind, we can proceed to cubes up to a billion. A three-digit cube root abc bears only one problem, which is finding the middle digit, b. For this, we will need to add another column to our table. We're going to use a divisibility rule. This is a simple rule which tells us when a number is divisible by another; that is, when it divides evenly into it without leaving any remainder. We will have to look at divisibility by eleven.

We calculate the alternating digit sum of a number by: taking the last digit, subtracting the 2nd last digit, adding the 3rd last digit, subtracting the next digit and so on, always taking the digits from the right If we get an answer which is below 0, we add 11 repeatedly until the answer is in the range 0 - 10. If we get an answer which is 11 or greater, we subtract 11 repeatedly until the answer is in the range 0 - 10. We're not going to prove it here, but the alternating digit sum gives us the remainder after the original number is divided by 11. An example would be 184,758: its alternating digit sum equals 8-5+7-4+8-1 = 13. We take 11 from this to bring it into the range 0 - 10, giving us 2. This means that 184,758 is not divisible by 11, because it has a remainder of 2 when divided by 11.

The following table is the original one with an additional column: the remainder when the cube is divided by 11. 0 1,000 10 If you are given a cube between a million and a billion, determine its remainder when divided by eleven, by calculating its alternating digit sum. Then look up in the memorised table which remainder its root has: find the alternating digit sum in the third column, and read the remainder from the first column. For an alternating digit sum of two it would be seven, for example, or for eight it would be two. Do not mix up the columns! With the remainder r and the easily computed first and last digit you can get b like this: b = a + c - r That is, add the first and second digit, then subtract the remainder to get the middle digit. Example Find the cube root of 122,023,936. The cube root's digits are determined as follows: Our cube is more than 64,000,000 but less than 125,000,000, so the first digit of the cube root must be 4.

The last digit of the cube is 6 as in 216, so the last digit of the cube root must also be 6. The alternating digit sum equals 6-3+9-3+2-0+2-2+1, which is 12. Subtract 11 from this to get it into the range 0 - 10, giving us 1. 1 in the third column corresponds to 1 in the first column, so r is 1. Add the first and third digits and subtract r to get 4+6-1 = 9, the second digit. So the cube root must be 496 and we have 4963=122,023,936. After a little practice, you will soon be the centre of every party . Advanced Math: Calculating 23rd Root of a 201 Digit Whole Random Number., reply thanks, scientific calculator Calculating 23rd Root of a 201 Digit Whole Random Number. Advanced Math/Calculating 23rd Root of a 201 Digit Whole Random Number.

Advertisement I was trying to trace the algorithm/logic/method for Calculating 23rd Root of a 201 Digit Whole Number. 1. Number 9 is a single digit number, Number 99 is a two digit number, Number 999 is a three digit number and so on .... 2. First find the square root, cube root, 4th root, 5th root, till 23rd root of 9 with help of scientific calculator / computer 3. Similarly find the square root, cube root, 4th root, 5th root, till 23rd root of 99, 999, 9999, 99999, 999999, 9999999 till the number 9 suffix with number 9 - 201 times with help of scientific calculator / computer. 4. This will give us 23root of a 201 digit number 9 suffix with number 9 - 201 times. But to calculate a 23rd root of any random number will be some other method ? Will the above method/logic help to some extend ?.

Awaiting your reply, Prashant S Akerkar Answer I'm afraid I don't understand your algorithm with the numbers made up of multiple 9s. If I was to find the 23rd root of a huge number N I would first calculate y = log(N^1/23) = (1/23)log(N) and then get 10^y = N^(1/23). I also note that log10(201) = 2.3, so maybe you are thinking of your number 23 = A^2 as an intensity (A = amplitude) and finding the dB = 20log10(A). Why you would do this I have no idea! Questioner's Rating

Question: What single digit whole number is the cube root of 64?
Answer: 4