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 $x$ and there are $n$ numbers, this is the following sum:

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