Mental Math: Cube-Roots and More
\[ \definecolor{modcolor}{RGB}{128,192,255} \newcommand{\clrmod}[1]{ { \color{modcolor} \mod{#1} } } \newcommand{\modelev}[]{\clrmod{11}} \]2-Digit Cube-Roots (Very Easy)
Taking an exact 2-digit cube-root in your head is actually very easy. For example, given \(571 787\), you should be able to instantly say that it is \(83^3 = 83 \times 83 \times 83\). If you don't know the trick, this feat seems impossible. But it's actually so simple a young child could do it!
Learn the following table of 1-digit cubes:
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) | \(10\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(x^3\) | \(0\) | \(1\) | \(8\) | \(27\) | \(64\) | \(125\) | \(216\) | \(343\) | \(512\) | \(729\) | \(1000\) |
Now, here's how the trick works.
The first digit of the answer, you get by comparing the thousands to the table. In our example, the thousands of \(571 787\) is \(571\). Comparing to entries in the table, we see that \(571\) is between[1] \(512\) and \(729\). These are the entries for \(8^3\) and \(9^3\), so we can conclude that the cube-root is between \(80\) and \(90\). That is, the first digit is \(8\).
The last digit of the answer, you get by finding the last digit of the cube in the table. The last digit of \(571 787\) is \(7\), and looking in the table we see that only \(3^3=27\) ends in \(7\), so the last digit is \(3\).
Thus, the final answer is \(83\) and we are done!
Let's try another example: \(\sqrt[3]{74 088}\). The thousands is \(74\), which is between \(64\) and \(125\), so the first digit is \(4\). The cube ends in \(8\), and looking in the table, we see that only \(2^3=8\) ends in \(8\), so the last digit is \(2\). The answer is \(42\).
It really is that simple!
Here are some exercises to practice:
- \(\sqrt[3]{493 039} =\)
\(~~~79~~~\) - \(\sqrt[3]{~~~~9 261} =\)
\(~~~21~~~\) - \(\sqrt[3]{~~15 625} =\)
\(~~~25~~~\) - \(\sqrt[3]{287 496} =\)
\(~~~66~~~\) - \(\sqrt[3]{729 000} =\)
\(~~~90~~~\)
Please note that this only works when the answer is known to be an exact integer. The cube-root of \(1331\) is correctly calculated to be \(11\), but the cube-root of \(1332\) is not \(18\) (it is actually \(11.00275\cdots\)).
3-Digit Cube-Roots (Harder)
It's possible to expand the technique for exact 3-digit cube-roots as well.
For a 3-digit cube-root, we find the first digit the same way as before (except using the millions instead of the thousands). We also find the last digit the same way as before.
For the middle digit, we have to first add another line of intermediary data to our table:
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) | \(10\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(x^3\) | \(0\) | \(1\) | \(8\) | \(27\) | \(64\) | \(125\) | \(216\) | \(343\) | \(512\) | \(729\) | \(1000\) |
| \(x^3~\modelev\) | \(0\) | \(1\) | \(8\) | \(5\) | \(9\) | \(4\) | \(7\) | \(2\) | \(6\) | \(3\) | \(10\) |
The middle digit is the root's first digit, plus its last digit, minus the modular cube-root of the cube modulo \(11\), all modulo \(11\). (Note: if you haven't encountered it before, 'modulo \(11\)' just means 'remainder after you divide by \(11\)'. See for more.)
Whew! That's quite a mouthful! Let's unpack it with an example.
Suppose we have \(\sqrt[3]{101 847 563}\). Using the millions (\(101\)), we see that the first digit of the cube-root is \(4\). Using the last digit, \(3\), we see that the last digit of the cube-root is \(7\).
We now know that the number is of the form '\(4\_7\)'. For the middle digit:
- Calculate the modulus of the cube: \[101 847 563 \equiv 4~\modelev\]
- Find the result in the last row of the table and then get the modular cube-root (first row) that corresponds to it: \[4 \rightarrow 5\]
- Subtract this from the sum of the first and last digits. Then take the modulus again to get the middle digit: \[(4+7)-5 \equiv 6~\modelev\]
Thus, the final answer is \(467\).
Pay careful attention to the order of these steps—do not mix them up![2]
The difficult part is computing the modulo \(11\) of the cube.
Doing this using division is doable in your head, but that's not the best way. The best way is to use a trick[3][4]: starting from the rightmost digit of the cube, alternately add and subtract digits. For our example:
\begin{align*} &101 847 563 \equiv 4~\modelev\\ &+\!3\,-\!6\,+\!5\,-\!7\,+\!4\,-\!8\,+\!1\,-\!0\,+\!1 \equiv 4~\modelev \end{align*}Personally, I find it less error-prone to separate the digits into those that get added versus those that get subtracted, and then subtract the groups:
\[ (3+5+4+1+1) - (6+7+8+0) \equiv 4~\modelev \]Let's do another example, \(\sqrt[3]{30 959 144}\). Using the millions \(30\), we see that the first digit of the cube-root is \(3\). Using the last digit, \(4\), we see that the last digit of the cube-root is \(4\).
The cube's modulus is computed by summing digits: \(+\! 4 \,-\! 4 \,+\! 1 \,-\! 9 \,+\! 5 \,-\! 9 \,+\! 0 \,-\! 3 \equiv 7~\modelev\). We then look this up in the table to find that the modular cube-root of \(7\) is \(6\). We then subtract this from the other two digits: \((3+4)-6 \equiv 1~\modelev\), which is the middle digit.
The final answer is \(314\).
Practice problems:
- \(\sqrt[3]{217 081 801} =\)
\(~~~601~~~\) - \(\sqrt[3]{~~63 044 792} =\)
\(~~~398~~~\) - \(\sqrt[3]{~~~~1 860 867} =\)
\(~~~123~~~\) - \(\sqrt[3]{~~94 818 816} =\)
\(~~~456~~~\) - \(\sqrt[3]{997 002 999} =\)
\(~~~999~~~\)
Other Roots
Cube-roots are useful, but square-roots are even more useful. Why not have a trick for that?
Unfortunately, the basic answer is that the powers for squares do not have unique last digits, so you cannot find the last digit unambiguously.
Here's the table for some other powers:
| \(x\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) | \(10\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(x^2\) | \(0\) | \(1\) | \(4\) | \(9\) | \(16\) | \(25\) | \(36\) | \(49\) | \(64\) | \(81\) | \(100\) |
| \(x^3\) | \(0\) | \(1\) | \(8\) | \(27\) | \(64\) | \(125\) | \(216\) | \(343\) | \(512\) | \(729\) | \(1 000\) |
| \(x^4\) | \(0\) | \(1\) | \(16\) | \(81\) | \(256\) | \(625\) | \(1 296\) | \(2 401\) | \(4 096\) | \(6 561\) | \(10 000\) |
| \(x^5\) | \(0\) | \(1\) | \(32\) | \(243\) | \(1 024\) | \(3 125\) | \(7 776\) | \(16 807\) | \(32 768\) | \(59 049\) | \(100 000\) |
| \(x^6\) | \(0\) | \(1\) | \(64\) | \(729\) | \(4 096\) | \(15 625\) | \(46 656\) | \(117 649\) | \(262 144\) | \(531 441\) | \(1 000 000\) |
| \(x^7\) | \(0\) | \(1\) | \(128\) | \(2 187\) | \(16 384\) | \(78 125\) | \(279 936\) | \(823 543\) | \(2 097 152\) | \(4 782 969\) | \(10 000 000\) |
Notice how for \(x^2\), we have that \(4^2=16\) and \(6^2=36\), which both end in \(6\). If you think about it, for any positive even power, the columns for \(4\) and \(6\) will end in \(6\): each additional power of \(2\) just multiplies that last digit by \(6\), resulting in another \(6\). So all positive even powers are ambiguous!
Thus, we cannot use this table-based approach for square-roots (or 4th-roots or 6th-roots or . . .). World-class human calculators use an approach based on estimation and large memorized tables, but I find Newton iteration more practical.
I've omitted powers larger than \(7\), because the numbers just keep getting bigger and more unwieldy, while the likely usefulness diminishes. However, returning to the odd powers in the table, we can see that 5th-roots and 7th-roots will also work with an analogous 2-digit method—and the 5th-root is particularly nice because the last digits are in order!
For example, consider the fifth-root \(\sqrt[5]{69 343 957}\). Because it is a 5th-root instead of a 3rd-root, we need to look at the most and least significant groups of five digits, not three digits. Other than that, the 2-digit algorithm proceeds similarly as for cubes.
The digits five places above are \(693\). This is between \(243\) and \(1024\), so the first digit is \(3\). Similarly, the last digit in the power is \(7\), so the last digit in the root is \(7\). The answer is \(37\).
For example, consider the seventh-root \(\sqrt[7]{17 565 568 854 912}\). Because it is a 7th-root instead of a 3rd-root, we need to look at the most and least significant groups of seven digits, not three digits. Other than that, the 2-digit algorithm proceeds similarly as for cubes.
The digits seven places above are \(1 756 556\). This is between \(823 543\) and \(2 097 152\), so the first digit is \(7\). Similarly, the last digit in the power is \(2\), so the last digit in the root is \(8\). The answer is \(78\).