Not sure where I originally saw this, but it had to be at least 20 years ago, and I thought it was pretty cool and seems to fit into this category.

Even seen someone compute cube roots quickly? With a little practice, you can have someone cube any 2 digit number, tell you the cube, and as quickly as they say the cube, you can tell them the cube root.

All you have to do is remember the cubes of the single digit numbers:

0^3 = 0
1^3 = 1
2^3 = 8
3^3 = 27
4^3 = 64
5^3 = 125
6^3 = 216
7^3 = 343
8^3 = 512
9^3 = 729

Notice that for 2,3,7,8, the last digit of the cube is 10 minus the original number, and for the other digits, the last digit of the cube is the same as the original number. (All I ever really remembered was 2 and 3, then 10-2 and 10-3, for which numbers were reversed.)

Now, when someone tells you a number, just listen for the digits left of the comma and for the final digit. The first digit of the cube root is the next lowest perfect cube in the list above, and the second digit is based on the formula in the preceding paragraph.

All of this is easier to show than describe. Let's look at a couple of examples:

Suppose you get 373,248. Well, 343 is the largest perfect cube less than 373 (the digits left of the comma), so the first digit is 7. Using the 2,3,7,8 tule above, the second digit must be 2. So... 72^3 = 373, 248.

Look at 274,625. 274 > 216, so the first digit is 6. Second digit is 5 from the rule, so 65^3 = 274, 625.

Cool, huh? Or maybe it's just the math geek in me. I can remember impressing my friends in calculus class in high school with this. Haven't really used it since then for some reason.

Eric.

If you like cube roots, then you'll love fifth roots - they're even simpler.

They're simpler, that is, as long as you practice!

You can do this kind of maths by memorizing some logarithms. For example draw 2-digit-roots from 6-digit-numbers...

I did that some 20 years ago, cannot remember it exactly, but I think I can figure it out again. Anyone interested?

Heinz

And there is much more you can do with logarithms as you can imagine...

Heinz