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Difference between revisions of "2016 AMC 10B Problems/Problem 18"

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<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7</math>
 
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7</math>
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==Solution==
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Let us have two cases, where <math>345</math> is the sum of increasing odd number of numbers and even number of numbers.
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Case 1: Sum of increasing odd number of numbers:
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The mean of an arithmetic sequences with an odd number of numbers is the middle term.
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Let us call the middle term <math>x</math>, and the number of terms <math>n</math>.
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                                <math>x*n=345</math>
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We can break down <math>345</math> into <math>3*5*23</math>. Thus our possible <math>(x,n)</math> are the following: <math>(1,345)</math>,<math>(3,105)</math>,<math>(5,69)</math>,<math>(15,23)</math>,<math>(23,15),</math>(69,5),<math>(105,3)</math>,<math>(345,1)</math>

Revision as of 12:58, 21 February 2016

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),$(105,3)$,$(345,1)$

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