Ask a friend to think of a two-digit number, cube it, and tell you the answer. Then you can tell them the cube root straight off, without having to do any calculation.

How? Look at the last digit of the number they’ve given you. If they’ve given you 250047, for example, that’s 7.

The second digit of the cube root is the same, except that 2 swaps with 8 and 3 swaps with 7. If the last digit is 2 then the second digit of the root is 8; if the last digit is 3 then second digit of the root is 7; and vice versa.

So for 250047, last digit is 7, second digit of root is 3.

Then forget about the last three digits of the number you’re given, and just look at the first ones: 250~~047~~ gives you 250 in this case.

The first digit of the root is the biggest digit with a cube smaller than what you’ve now got. In this case, 6, because 6^{3}=216, but 7^{3}=343, which is bigger than 250.

In this example, cube root of 250047 is 63.

Your challenge:

1. Why isn’t there a similar easy way to find square roots?

2. What other roots is there a similar easy method for (with two-digit numbers)? 4th roots? 5th roots? 6thâ€¦?

Joan Onokhua and Hamse Adam found solutions, and the Year 12 Further Maths class also found the solution collectively, with Jetmir Guri providing most of the insights.

Answer: in “clock arithmetic” with ten hours on the clock, numbered 0 to 9:

1^{3} = 1

2^{3} = 8

3^{3} = 7

4^{3} = 4

5^{3} = 5

6^{3} = 6

7^{3} = 3

8^{3} = 2

9^{3} = 9

so in “clock arithmetic” with ten hours on the clock, every number has a unique cube root. In other words, when the cube is written out in ordinary arithmetic, its last digit (which equals its value in “clock arithmetic” with ten hours on the clock) tells you what number was cubed.

However, again in “clock arithmetic” with ten hours on the clock

2^{2} = 4 and also

8^{2} = 4

so you *can’t* tell from the last digit what number was squared.

This happens because

a^{2} = b^{2} in “clock arithmetic” with ten hours on the clock

when

a^{2} − b^{2} = 0 in that “clock arithmetic”

in other words when a=b *or* when a+b=0 (example:2+8=0 in “clock arithmetic” with ten hours on the clock).

However, (a+b) does *not* divide exactly into a^{3} − b^{3}

The trick works for all *odd* powers, and for no even powers.