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.

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

To find the solution, we need to follow these steps:

1. Determine the factors of 2008.
2. Determine the number of factors of 2008.
3. Find the sum of consecutive integers with a number of terms twice the number of factors of 2008.
4. Identify three consecutive integers in the sum.

Let's start by finding the factors of 2008:
The prime factorization of 2008 is 2^3 * 251.

To find the factors, we can use the prime factorization. The factors of 2008 will be generated by multiplying different combinations of the prime factors 2 and 251:
1, 2, 4, 8, 251, 502, 1004, 2008

Now, let's determine the number of factors of 2008:
From the prime factorization, we have 3 exponents for 2 and 1 exponent for 251.
(3 + 1) * (1 + 1) = 8 factors in total.

Next, we need to find the sum of consecutive integers with a number of terms twice the number of factors of 2008:
Since there are 8 factors, we need to find the sum of 16 consecutive integers.

To find the sum of consecutive integers, we can use the formula: sum = (first term + last term) * number of terms / 2.

Let's denote the first term as 'x', so the 16th term will be 'x + 15'.
Plugging these values into the formula, we have:
sum = (x + (x + 15)) * 16 / 2
sum = (2x + 15) * 8

Now, we have the equation: (2x + 15) * 8 = 2008.

Let's solve this equation for 'x':
2x + 15 = 2008 / 8
2x + 15 = 251
2x = 251 - 15
2x = 236
x = 236 / 2
x = 118

Now, we have found the first term of the consecutive integers as 118.

Finally, we need to identify the smallest of three consecutive integers:
So the three consecutive integers are 118, 119, and 120. The smallest of these three is 118.

Therefore, the smallest of the three consecutive integers is 118.