TMUA practice: Arithmetic Series, problem 5

Let the first term be the integer $a$. With common difference $2$, the sum of the first $n$ terms is

$S_{n}=\frac{n}{2}\bigl[2a+(n-1)\cdot2\bigr]=n\,(a+n-1).$

Setting $S_{n}=153$ gives $n\,(a+n-1)=153$.

Which $n$ are possible? Since $a+n-1$ must be an integer, $n$ must be a divisor of $153$. Conversely, for any divisor $n$ the value $a=\dfrac{153}{n}-n+1$ is an integer, and $a$ is allowed to be any integer (positive, negative or zero). So every divisor $n>1$ works.

Count the divisors. $153=3^{2}\cdot17$, whose divisors are $1,3,9,17,51,153$. Those with $n>1$ are $3,9,17,51,153$ — five values.

Answer: D.