Question

Given that the product of 22 integers is equal to 1, prove that their sum is not equal to 0.

Answer 1

Fun problem.

We have a positive product, so no zero factors.

We have a positive product, so an even number of negative factors.

We have a product of 1, so all the integers must be either +1 or -1.

If we have 2n factors of -1 we have 22-2n factors of 1 so a sum of

$s = (-1)2n + (1)(22-2n) = 22-4n$

For that to be zero we need

$0 = 22-4n$

$n = 22/4 = 11/2$

That's not an integer; we've shown there's no even number of -1s that will make the sum zero, and since there must be an even number of -1s, the sum cannot equal zero.

Answer 2

1. If the product of these integers is to be 1, then all of them must be either 1 or -1.

2. Since the product is positive (+1), it must be that there are an *even* number of negative ones (-1), if any.

3. If the sum were 0 it would mean that the number of +1's must equal the number of -1's. So that means there would have to be exactly 22/2=11 of each.

4. But if there were 11 of each, that means the number of -1's would be *odd* and there's no way the product could be +1 (as stated in 2 above).

Hence, the sum is never 0, if the product of 22 integers is equal +1.