Asked by tiffani
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.
Answers
Answered by
Steve
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
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
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