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

Revision as of 13:37, 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

Factorize $345=3\cdot 5\cdot 23$ .

Suppose we take an odd number $k$ of consecutive integers, centred on $m$ . Then $mk=345$ with $\tfrac12k<m$ . Looking at the factors of $345$ , the possible values of $k$ are $3,5,15,23$ centred on $115,69,23,15$ respectively.

Suppose instead we take an even number $2k$ of consecutive integers, centred on $m$ and $m+1$ . Then $k(2m+1)=345$ with $k\le m$ . Looking again at the factors of $345$ , the possible values of $k$ are $1,3,5$ centred on $(172,173),(57,58),(34,35)$ respectively.

Thus the answer is $\textbf{(E) }7$ .

2016 AMC 10B (Problems • Answer Key • Resources)
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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.
+Factorize <math>345=3\cdot 5\cdot 23</math>.
-Case 1: Sum of increasing odd number of numbers:
+Suppose we take an odd number <math>k</math> of consecutive integers, centred on <math>m</math>. Then <math>mk=345</math> with <math>\tfrac12k<m</math>.
-The mean of an arithmetic sequences with an odd number of numbers is the middle term.
+Looking at the factors of <math>345</math>, the possible values of <math>k</math> are <math>3,5,15,23</math> centred on <math>115,69,23,15</math> respectively.
-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>,<math>(69,5)</math>,<math>(105,3)</math>,<math>(345,1)</math>
-However, the erroneous solutions which have negatives or have only 1 term are the following:
-<math>(1,345)</math>,<math>(3,105)</math>,<math>(5,69)</math>,<math>(15,23)</math>,<math>(345,1)</math>
-So, we are left with:
-<math>(23,15)</math>,<math>(69,5)</math> and <math>(105,3)</math>
-Which gives us three possible ways.
-Case 2: Sum of increasing even number of numbers:
+Suppose instead we take an even number <math>2k</math> of consecutive integers, centred on <math>m</math> and <math>m+1</math>. Then <math>k(2m+1)=345</math> with <math>k\le m</math>.
-If the first term is <math>x</math> and there are <math>n</math> numbers, this is the following sum:
+Looking again at the factors of <math>345</math>, the possible values of <math>k</math> are <math>1,3,5</math> centred on <math>(172,173),(57,58),(34,35)</math> respectively.
- <math>x+x+1+x+2+...+x+(n-1)=</math>
- <math>xn+1+2+3+...+n-1=</math>
- <math>xn+n(n-1)/2=</math>
- <math>n/2(2x+n-1)=</math>
-In this case we have our following <math>(x,n)</math>:
-<math>(-344,690)</math>,<math>(-108,210)</math>,<math>(-66,138)</math>,<math>(-15,46)</math>,<math>(-3,30)</math>,<math>(30,10)</math>,<math>(50,6)</math>,<math>(172,2)</math>
-Taking out our erroneous solutions:
-<math>(30,10)</math>,<math>(50,6)</math>,<math>(172,2)</math>
-Which also gives us three ways.
-Counting our cases: <math>3</math>+<math>3</math>=<math>6</math>,<math>D</math>
+Thus the answer is <math>\textbf{(E) }7</math>.
 ==See Also==
 {{AMC10 box|year=2016|ab=B|num-b=17|num-a=19}}
 {{MAA Notice}}

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

Revision as of 13:37, 21 February 2016

Problem

Solution

See Also