2016 AMC 10B Problems/Problem 18

Revision as of 12:12, 21 February 2016 by Shrummy (talk | contribs) (Solution)

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 $n$.

 $x*n=345$

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:

$x+x+1+x+2+...+x+(n-1)=$
$xn+1+2+3+...+n-1=$
$xn+n(n-1)/2=$
$n/2(2x+n-1)=$

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:$3$+$3$=$6$,$D$