I found this in a book...

1 + 6 + 7 + 17 + 18 + 23 = 2 + 3 + 11 + 13 + 21 + 22;
1^2 + 6^2 + 7^2 + 17^2 + 18^2 + 23^2 = 2^2 + 3^2 + 11^2 + 13^2 + 21^2 + 22^2;
1^3 + 6^3 + 7^3 + 17^3 + 18^3 + 23^3 = 2^3 + 3^3 + 11^3 + 13^3 + 21^3 + 22^3;
1^4 + 6^4 + 7^4 + 17^4 + 18^4 + 23^4 = 2^4 + 3^4 + 11^4 + 13^4 + 21^4 + 22^4;
1^5 + 6^5 + 7^5 + 17^5 + 18^5 + 23^5 = 2^5 + 3^5 + 11^5 + 13^5 + 21^5 + 22^5.

mathsyperson wrote:

What an amazing coincidence! It must be coincidence because it stops working when you use any power higher than 5.

It's a consequence of the following even bigger coincidence!

a^n + (a + 4b + c)^n + (a + b + 2c)^n + (a + 9b + 4c)^n + (a + 6b + 5c)^n + (a + 10b + 6c)^n = (a + b)^n + (a + c)^n + (a + 6b + 2c)^n + (a + 4b + 4c)^n + (a + 10b + 5c)^n + (a + 9b + 6c)^n,

where a, b, c are any positive integers and n can be 1, 2, 3, 4, or 5.

See also http://nrich.maths.org/askedNRICH/edited/412.html and http://www.primepuzzles.net/puzzles/puzz_065.htm

Last edited by NIH (2005-08-14 09:39:19)

Here's an interesting one:

111111111² = 12345678987654321.

Basically, when any number composed entirely of n ones is squared, the result will be 12...n...21. Well, at least when n < 10, once we get to 10 the middle gets screwy.

Last edited by Zhylliolom (2005-09-06 10:33:34)