## Taming The Sum

#### Problem

Show that the sum, 123...99=100, where between each term can be independently set to + or , has at least one solution.

#### Solution

Consider the sum, `n` (`n`+1) (`n`+2) + (`n`+3) = 0.

In other words, we can set four consecutive integers to zero:

1 2 3 + 4 = 0

5 6 7 + 8 = 0

9 10 11 + 12 = 0

...

93 94 95 + 96 = 0

5 6 7 + 8 = 0

9 10 11 + 12 = 0

...

93 94 95 + 96 = 0

Then -97 + 98 + 99 = 100

What about 1^{2}2^{2}3^{2}...99^{2} = 100?

Is there a solution for cubes?

Problem ID: 140 (Dec 2003) Difficulty: 2 Star