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== | ||
+ | Let us have two cases, where <math>345</math> 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 <math>x</math>, and the number of terms <math>n</math>. | ||
+ | <math>x*n=345</math> | ||
+ | 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 11:58, 21 February 2016
Problem
In how many ways can be written as the sum of an increasing sequence of two or more consecutive positive integers?
Solution
Let us have two cases, where 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 , and the number of terms .
We can break down into . Thus our possible are the following: ,,,,(69,5),,