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

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The mean of an arithmetic sequences with an odd number of numbers is the middle term.  
 
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>.
 
Let us call the middle term <math>x</math>, and the number of terms <math>n</math>.
−
                                <math>x*n=345</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>
 
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>
 +
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>(69,5)and <math>(105,3)</math>
 +
Which gives us three possible ways.
 +
Case 2: Sum of increasing even number of numbers:
 +
If the first term is <math>x</math> and there are <math>n</math> numbers, this is the following sum:
 +
<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 (x,n):
 +
<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>(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$

Revision as of 12:12, 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),$(105,3)$,$(345,1)$ 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:$3$+$3$=$6$,$D$

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