1973 AHSME Problems/Problem 21

Problem

The number of sets of two or more consecutive positive integers whose sum is 100 is

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution

If the first number of a group of $n$ consecutive numbers is $a$, the $n^\text{th}$ number is $a+n-1$. We know that the sum of the group of numbers is $100$, so \[\frac{n(2a+n-1)}{2} = 100\] \[2a+n-1=\frac{200}{n}\] \[2a = 1-n + \frac{200}{n}\] We know that $n$ and $a$ are positive integers, so we check values of $n$ that are a factor of $200$. Of these values, the only ones that result in a positive integer $a$ is when $n = 5$ or when $n = 8$, so there are $\boxed{\textbf{(B)}\ 2}$ sets of two or more consecutive positive integers whose sum is $100$.

See Also

1973 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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