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## Digging Up 3-Digit Cube Roots

Published on Sunday, November 17, 2013 in , , , , Over in the Mental Gym, you can learn how to mentally work out the cube root of any perfect cube up to 1,000,000!

It's time to take that feat to the next level, so in today's post, you'll learn how to mentally work out the cube roots of perfect cubes between 1 million and 1 billion!

BASICS: To learn this feat, you should already be comfortable with the techniques taught in the Mental Gym for determining cube roots of perfect cubes up to 1 million. There's also a Scam School video on working out 2-digit cube roots of perfect cubes which you may find helpful.

To start, you'll need a volunteer to choose the number, and they need to be using a calculator that can display up to 9 digits. Fortunately, most smartphone-based calculators can handle this.

Have the volunteer enter any whole 3-digit number into the calculator, and then ask them to multiply it by itself and by itself again. In other words, if they chose the number 123, they'd multiply 123 × 123 × 123. Once they've done that, they should read off the number from their calculator, and you then write it down. You write it down because you'll need to study it as you work through it.

FINDING THE HUNDREDS AND ONES DIGITS: Just as with the original 2-digit version, you start by mentally breaking up the digits into groups of 3s. It's often helpful to write the number given to you with commas, which visually break up the number into groups of 3 automatically.

As an example, let's say you're given the number 15,069,223. Written that way, you can already see 15 as the leftmost group, 069 as the middle group, and 223 as the rightmost group. The leftmost group can consist of either 1, 2, or 3 digits, while the other groups will always consist of 3 digits.

You get the hundreds digit in the same way as in the 2-digit version (mentioned above in BASICS). You look at the leftmost set, in our example number of 15,069,223, that would be 15. We see that 15 is greater than 23 (8) and less than 33 (27), so the leftmost digit of the cube root would be 2 in our example.

To get the ones digit, you look at the rightmost digit, and work out which number, when cubed, would result in that final digit. Again, this is discussed in the tutorials linked under BASICS. Going back to our 15,069,223 example, we see that the last digit is 3. The only number which gives a number ending in 3 when cubed is 7, so the ones digit of the answer must be 7.

At this point, we know 2 of the 3 digits in the answer. In our example, we've already determined that the cube root must be 2 hundred-something and 7, or 2_7.

FINDING THE TENS DIGIT: To find the middle digit, we need to consider the entire number in a very particular way. What we need to know is what the remainder would be when the entire number is divided by 11. This is easy with small numbers, such as 68. We know 66 is a multiple of 11, so it's easy to work out that 68 has a remainder of 2 when divided by 11, but how do you deal with numbers of 7-, 8-, or 9-digits?

Fortunately, there are simple shortcuts you can use, and yes, they're simple enough to do in your head.

The classic trick is to start with the rightmost digit of the given number (the ones digit, in other words), subtract the number to its left, add the number to the left of the next one, and so on, alternately adding and subtracting as you go.

Our example of 15,069,223 becomes 3 - 2 + 2 - 9 + 6 - 0 + 5 - 1, or 4. In your mind, you might think of it more as, #147;3 minus 2 is 1, plus 2 is 3, minus 9 is -6, plus 6 is 0, minus 0 is 0, plus 5 is 5, minus 1 is 4.” Note that, if we double check the problem of 15.069,223 mod 11 with Wolfram|Alpha, we also get 4. Working this way, you'll always get an answer from 0 to 10, but how do we use this number?

Note that the answer we get at this point is representative of the large number in question, not the 3-digit answer. Basically, we need to ask ourselves what type of number, when cubed and divided by 11, would leave the specific remainder (4, in our example). Here's how we do that.

Each number from 0 to 10, when cubed, leaves a unique remainder when divided by 11, as shown in this table on Wolfram|Alpha. This pattern repeats starting over again at each multiple of 11. So, any number 2 greater than a multiple of 11 when cubed and divided by 11, will always leave a remainder of 8. Any number 3 greater than a multiple of 11, when cubed and divided by 11, will always leave a remainder of 5, and so on.

Since we're starting from the cubed, number, we look for the remainder we've determined, and find the number above it on the table. Our example's remainder was 4, which can be found underneath the 5 in that table. This means that the cube root must be 5 greater than a multiple of 11.

Before I continue, let me discuss memorizing the table. It may seem hard, but it relly isn't. First, notice that the cubes which, when divided by 11, leave a remainder of 0, 1, or 10, will have a cube root that, when divided by 11, will leave a remainder of 0, 1, or 10 respectively. Those are the easiest ones to recall.

If you look at 23 mod 11, note that the result will be 8. Now look at 83 mod 11, which is 6. 63 mod 11 is 7, and 73 mod 11 is 2. Those 4 numbers make a small cycle of {2, 8, 6, 7}. When working from the large perfect cube, you find the remainder after dividing by 11, look on that list, and then look for the number immediately before it for the number you need to determined the root. 7 would give you 6, 6 would give you 8, 8 would give you 2, and 2 would give you 7 (note how this cycles around).

There's one more cycle to help you memorize the table. 33 mod 11 is 5, 53 mod 11 is 4, 43 mod 11 is 9, and 93 mod 11 is 3. This other cycle, then, is {3, 5, 4, 9}. Similar to the previous cycle, you find the remainder of the large cube when divided by 11, look on the list to the number immediately before it, and you'll get the number that will help you determined the root.

Since we determined that 15,069,223 mod 11 is 4 with a quick calculation, we can recall that 4 is part of the {3, 5, 4, 9} cycle, and that since the 5 is right before the 4, the root must be 5 greater than a multiple of 11. We did this earlier by looking at the table, but you want to be able to recall the table from memory when demonstrating this feat.

At this point, we know 2 important things about the cube root. Our example cube root looks like 2_7, and is 5 greater than a multiple of 11. That's enough to narrow down the answer to one specific number!

Remember the technique of alternately adding an subtracting digits to get the remainder after dividing by 11? We're going to use it again here. So, the example problem becomes 7 - x + 2 = 5. That simplifies to 9 - x = 5, and the answer is obviously 4, because 9 - 4 = 5. That means that 4 is the digit in the middle! Our example cube root, then, must be 247. Double-checking our answer with Wolfram|Alpha, we find that 247 is 15,069,223!

More generally, you can just add the two numbers you have and subtract the remainder number you worked out. We take 2_7 and add 2 + 7 to get 9, and then subtract the remainder, 5, to get 9 - 5 = 4. Since we're talking about working through this in your head, that's the easiest way to go about it.

QUICKER EXAMPLE: This seems long only because it's new to you, and the example was interspersed with the description of the technique. Let's try a more streamlined example, and say we're given the number 241,804,367. What's the cube root?

Looking the leftmost group, we see 241 is between 63 and 73, so the answer is going to be in the 600s. The rightmost digit is a 7, and only 33 ends in 7, so our answer is along the lines of 6_3.

What is the remainder when that number is divided by 11? It's 7 - 6 = 1, + 3 = 4, - 0 = 4, + 8 = 12, - 1 = 11, + 4 = 15, - 2 = 13. 13 is actually larger than 11, so we subtract 11 from 13 to get 2.

So, the large perfect cube, when divided by 11, leaves a remainder of 2. 2 is part of the easily-recalled {2, 8, 6, 7} cycle, and right before the 2 is 7, so the cube root must be 7 more than a multiple of 11.

One last simple calculation, 6 + 3 (the two outside numbers) = 9, and 9 minus 7 (the remainder after dividing the cube root by 11) = 2, so the cube root must be 623! Sure enough, Wolfram|Alpha confirms that 6233 is 241,804,367.

PRACTICE: You can practice your newfound skill by having Wolfram|Alpha generate random perfect cubes between 1 million and 1 billion here.

While practicing, you'll also learn to watch for pitfalls. For example, I once got the number 367,061,696. I worked out that the leftmost group was between 73 and 83, so the cube root must be in the 700s. The rightmost digit is 6, which means that the rightmost digit of the cube root is 6. So far, I have 7_6.

Working out the remainder when dividing by 11, I got 6 - 9 = -3, + 6 = 3, - 1 = 2, + 6 = 8, - 0 = 8, + 7 = 15, - 6 = 9, + 3 = 12, and since 12 is greater than 11, I worked out 12 - 11 = 1. Recall that a remainder of 1 when the cube is divided by 11 means that the cube root has a remainder of 1 divided by 11, as well.

So far, so good, but we're about to encounter a small problem. We add together the two outer digits, 7 + 6 = 13, and then subtract 1 (the remainder of our cube root) for a final result of . . . 12?!? Obviously, 12 can't be the middle digit, as we're only looking for single digit. When you get a number equal to 11 or more, always subtract 11, and keep doing so until you get a number from 0 to 10. 12 - 11 = 1, so 1 is actually the middle digit we're looking. Therefore, the cube root of 716, and Wolfram|Alpha verifies this is correct.

Have fun, practice, and go out and amaze your friends and family with your newfound skill! I'd love to hear about any memorable reactions in the comments!