2016 AMC 10B Problems/Problem 18

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

Let us have two cases, where $345$ is the sum of increasing odd number of numbers and even number of numbers.

Case 1: Sum of increasing odd number of numbers: The mean of an arithmetic sequences with an odd number of numbers is the middle term. Let us call the middle term $x$ , and the number of terms $n$ .

We can break down $345$ into $3*5*23$ . Thus our possible $(x,n)$ are the following: $(1,345)$ , $(3,105)$ , $(5,69)$ , $(15,23)$ , $(23,15),$ (69,5) $,$ (105,3) $,$ (345,1) $However, the erroneous solutions which have negatives or have only 1 term are the following:$ (1,345) $,$ (3,105) $,$ (5,69) $,$ (15,23) $,$ (345,1) $So, we are left with:$ (23,15), $(69,5)and$ (105,3)$Which gives us three possible ways.

Case 2: Sum of increasing even number of numbers: If the first term is$ (Error compiling LaTeX. Unknown error_msg)x $and there are$ n $numbers, this is the following sum:$ x+x+1+x+2+...+x+(n-1)=$$ (Error compiling LaTeX. Unknown error_msg)xn+1+2+3+...+n-1=$$ (Error compiling LaTeX. Unknown error_msg)xn+n(n-1)/2=$$ (Error compiling LaTeX. Unknown error_msg)n/2(2x+n-1)= $In this case we have our following (x,n):$ (-344,690) $,$ (-108,210) $,$ (-66,138) $,$ (-15,46) $,$ (-3,30) $,$ (30,10) $,$ (50,6) $,$ (172,2) $Taking out our erroneous solutions:$ (30,10) $,$ (50,6) $,$ (172,2) $Which also gives us three ways. Counting our cases:$ 3 $+$ 3 $=$ 6 $,$ D$

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2016 AMC 10B Problems/Problem 18

Problem

Solution