Sum some numbers

Suppose $n$ is the smallest number.

Then $n + (n+1) + (n+2) + .... = 164$

Rewritten, where $m$ is the number of digits in the series:

$mn + (1+2+...+(m-1)) = 164$

$mn + m(m-1)/2 = 164$

$n = \frac{164 - \frac{m(m-1)}{2}}{m}$

$n = \frac{164}{m} - \frac{m-1}{2}$

We only have to consider $m = 2$ to $18$ since larger values will lead to a negative value for $n$ . And $m$ can't equal $1$ since the problem asks for "integers".

If $m$ is odd, then the right side of the equation $\frac{m-1}{2}$ will be an integer, however $164$ doesn't divide evenly by any odd number in the range. ( $41 > 18$ so $41$ is out of the range) in order for the left side to also be an integer.

Therefore, $m$ must be an even number between $2$ and $18$ , so the second term will be an integer $+ 0.5$ .

So, the first term needs to also be an integer $+ 0.5$ . This implies that $m$ can't have any prime factors other than $2$ .

That leaves $2,4,8,$ and $16$ . Of these, only $8$ gives us the required integer $+ 0.5$ for the first term.

The only numbers for which $n$ is a positive integer are $164$ when $m=1$ and $17$ when $m=8$ .

Therefore, since the problem states "integers" then $n = 17$ .