# Difference between revisions of "1964 AHSME Problems/Problem 28"

## Problem

The sum of $n$ terms of an arithmetic progression is $153$, and the common difference is $2$. If the first term is an integer, and $n>1$, then the number of possible values for $n$ is:

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 6$

## Solution

Let the progression start at $a$, have common difference $2$, and end at $a + 2(n-1)$.

The average term is $\frac{a + (a + 2(n-1))}{2}$, or $a + n - 1$. Since the number of terms is $n$, and the sum of the terms is $153$, we have:

$n(a+n-1) = 153$

Since $n$ is a positive integer, it must be a factor of $153$. This means $n = 1, 3, 9, 17, 51, 153$ are the only possibilities. We are given $n>1$, leaving the other five factors.

We now must check if $a$ is an integer. We have $a = \frac{153}{n} + 1 - n$. If $n$ is a factor of $153$, then $\frac{153}{n}$ will be an integer. Adding $1-n$ wil keep it an integer.

Thus, there are $5$ possible values for $n$, which is answer $\boxed{\textbf{(D)}}$.

## See Also

 1964 AHSC (Problems • Answer Key • Resources) Preceded byProblem 27 Followed byProblem 29 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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