Extra Credit Problem #1.

Solution:

Let x be the given number. Let n = (x-5)/10. So in the example x = 715 and n = 71.

We need to show that ( n(n+1) * 100  ) + 25 = x^2. . (Here x^2 means x squared and * means multiplied by).

Well, x^2 = (10n +5)* (10n + 5) = 100 n^2 + 100n + 25.

Also observe that ( n(n+1)* 100 ) + 25 = 100n^2 +100n +25. Therefore the two

quantities are equal.

QED