the number 2008 can be written as a sum of consecutive integers such that the number of terms in the sum is twice the number of factors of 2008. Find the smallest of three consecutive integers.

Answer 1

well, there's no way that 2008 can be the sum of three consecutive integers. I think you must have meant "these" not "three".

Otherwise, since 2008 has 8 divisors, we want 16 consecutive integers, meaning

16n + 15*16/2 = 2008
n = 118

So, the numbers are 118,...,133