# Difference between revisions of "2016 AMC 12B Problems/Problem 16"

## Problem

In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers? $\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

## Solution

We proceed with this problem by considering two cases, when: \par 1. There are an odd number of consecutive numbers \par 2. There are an even number of consecutive numbers

For the first case, we can cleverly choose the convenient form of our sequence to be $$a-n,\cdots, a-1, a, a+1, \cdots, a+n$$

because then our sum will just be $(2n+1)a$. We now have $$(2n+1)a = 345$$ and $a$ will have a solution when $\frac{345}{2n+1}$ is an integer, namely when $2n+1$ is a divisor of 345. We check that $$2n+1 = 3, 5, 15, 23$$ work, and no more, because $2n+1$ does not satisfy the requirements of two or more consecutive integers, and when $2n+1$ equals the next biggest factor, $69$, there must be negative integers in the sequence. Our solutions are $\{114,115, 116\}, \{67, \cdots, 71\}, \{16, \cdots, 30\}, \{4, \cdots, 26\}$.

For the even cases, we choose our sequence to be of the form: $$a-(n-1), \cdots, a, a+1, \cdots, a+n$$ so the sum is $\frac{(2n)(2a+1)}{2} = n(2a+1)$. In this case, we find our solutions to be $\{172, 173\}, \{55,\cdots, 60\}, \{30,\cdots, 39\}$.

We have found all 7 solutions and our answer is $\boxed{\textbf{(E)} \, 7}$.

## See Also

 2016 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 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 All AMC 12 Problems and Solutions

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