The method for extracting 3-digit cube roots of perfect cubes taught in my previous post is the classic approach for this task, but it's not the ONLY approach.

In this post, we'll look at some other ways to tackle the cube root challenge.

**QUICK TIP:** If you get a perfect cube ending in 3 zeroes, such as 238,328,000, after you put down the zero for the rightmost digit, you can ignore the zeroes in the cube, and work out the rest in the same way as you do for 2-digit cubes.

Working through our 238,328,000 example number, you can instantly put 0 down as the rightmost digit, and then think of the number as being 238,328 (effectively dividing it by 1,000). As in the standard method, you can see that 238 is greater than 6^{3} and less than 7^{3}, so the leftmost digit will be 6, giving us 6_0 so far. Now, the rightmost digit of our modified number, the final 8, indicates that 2 is the cube root, which gives us 620 as the answer. Sure enough, 620^{3} is 238,328,000!

**ALTERNATE MOD 11 APPROACH:** In the previous post, the approach for working out mod 11 involved alternately adding and subtracting number from right to left, and could result in numbers of 11 or more, requiring further adjustments. Here's another method you may prefer, as it's more straightforward.

To help you understand the process, we're going to go back to the technique of long division. Imagine you're asked to work out 94,818,816 ÷ 11 by long division, and give the remainder, as well. Writing it out, it would look something like this:

Consider that you'll have the perfect cube, 94,818,816 in our example above, written down already. You know mod 11 is important from the beginning, so there's no real need to write that down. Finally, we're *ONLY* interested in the remainder, so there's no need to keep track of a huge number such as 8619892 at all.

All you really need to do in this approach, then, is do the subtractions mentally in the same way, starting with the leftmost 2 digits (if the leftmost 2 digits are 10, then start with the leftmost 3 digits), and appending the next single digit to the right after each subtraction. In the example above, you'd look at the numbers on the paper as you go through them mentally, like this: “94 - 88 = 6...68 - 66 = 2...21 - 11 = 10...108 - 99 = 9...98 - 88 = 10...101 - 99 = 2...26 - 22 = 4” That 4 means that the perfect cube you were given, when calculated mod 11, is 4.

From that point, you would work through the procedure as taught in the previous post. The nice thing about this approach is that it will work for ANY modulus number for the same reason that long division always works. Once you get the hang of doing the subtractions and adding the next digit to the right, it's not hard at all.

**MOD 9 APPROACH:** There's also a mod 9 approach for finding the middle digit. Many people, including Arthur Benjamin, whose videos were recently discussed here and here, prefer this approach.

This version works on the idea of digital roots, also known as digit sums. Digital roots are a very simple idea, but if you need a refresher course, this video on digit sums and casting out 9s gives a good brief introduction to the idea, and this second video gives a little more detail.

Once you've got the leftmost and rightmost digits of the cube root, you're ready to use this technique. The first step is simply to add up all the digits of the perfect cube. If the total has 2 or more digits, you add up the individual digits of that number, and keep repeating this process until you get a 1-digit number, as described in the videos linked above.

For example, let's say you're given the perfect cube of 143,055,667. At this stage, you should already know that the cube root looks something like 5_3. You'd add 1 + 4 + 3 + 0 + 5 + 5 + 6 + 6 + 7 = 37. 37 has 2 digits, so you'd add 3 + 7 = 10. 10 is a 2-digit number, so you'd add 1 + 0 = 1, so the digital root of the perfect cube is 1.

How do we use this knowledge? Here, Wolfram Alpha shows us what happens when the number from 0 to 8 are cubed and then reduced mod 9 (which is the same as taking the digital root). Any number, when cubed, will always have a digital root of either 0, 1, or 8.

If a perfect cube has a digital root of 0 (or 9, which is the same thing), then its cube root must have a digital root of either 0 (or 9), 3, or 6. If a perfect cube has a digital root of 1, then its cube root must have a digital root of either 1, 4, or 7. And if a perfect cube has a digital root of 8, then its cube root must have a digital root of either 2, 5, or 8.

This table is definitely simpler to memorize than the mod 11 version, but it does come with a price. How do you know which cube root? Fortunately, a little rough judgement is all that is needed.

In our 143,055,667 example from above, we know that the cube root looks something like 5_3, and has a digital root of 1. This means that the cube root must have a digit root of 1, 4, or 7. Let's try the various possibilities and see what happens. 5 + 3 = 8, so we could get a digital root of 1 by adding 2 in the middle (since 5 + 2 + 3 = 10, and 10 has a digital root of 1). We could also get a digital root of 4 by putting a 5 in the middle (5 + 5 + 3 = 13, and 13 has a digital root of 4), or a digital root of 7 by putting an 8 in the middle (5 + 8 + 3 = 16, and 16 mod 9 is 7).

Now we have 3 possible answers: 523, 553, and 583. How do we choose? Go back and look at the leftmost group of numbers. Is the answer closer to 500 cubed (125 million), 600 cubed (216 million), or somewhere in the middle? If it were in the middle, the number would begin with a number around 170 million, and if it were towards the high end, the number would be close to 216 million, which rules out these possibilities. 143 million is much closer to 125 million, so we must be looking for the lowest of the 3 number, which is 523.

Double checking with Wolfram Alpha, sure enough, we find that 523^{3} = 143,055,667!

Dr. Benjamin has his own paper on finding cube roots here (PDF), where he discusses the mod 9 approach in more detail, if you're interested.

Consider these tricks and tips, put together your own preferred version of this feat, and then go out and amaze your friends and family with your newfound skills!

## Digging Up 3-Digit Cube Roots (Tips & Tricks)

Published on Thursday, November 21, 2013 in fun, math, mental math, self improvement, site features

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## 1 Response to Digging Up 3-Digit Cube Roots (Tips & Tricks)

This is a favorite feat of mine, as well as the late Devi Shakuntala. Integer roots follow all sorts of patterns that make solving them a lot easier than most people think. If you happen to know the cubes of all two digit numbers then this feat can be extended to cubes of 12 digits. You estimate the first two digits, check the last digit, then use modular arithmetic for the third digit. Mr. Cram's explanation could be extended to fifth, seventh, and even thirteenth roots as well (This would give a number of 39 digits!). Integer roots are very impressive for the little work they take to solve. Maybe in a future post Mr. Cram could explain how to extend this feat to other roots. It would make a great extension to the exponential expressions feat he already explained.

Keep up the great work! --Jay

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