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

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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>,</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>,<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:
 
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>
+
<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:
 
So, we are left with:
−
</math>(23,15),<math>(69,5)and </math>(105,3)<math>
+
<math>(23,15),</math>(69,5)and <math>(105,3)</math>
 
Which gives us three possible ways.
 
Which gives us three possible ways.
  
 
Case 2: Sum of increasing even number of numbers:
 
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:
+
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>x+x+1+x+2+...+x+(n-1)=</math>
−
  </math>xn+1+2+3+...+n-1=<math>
+
  <math>xn+1+2+3+...+n-1=</math>
−
  </math>xn+n(n-1)/2=<math>
+
  <math>xn+n(n-1)/2=</math>
−
  </math>n/2(2x+n-1)=<math>
+
  <math>n/2(2x+n-1)=</math>
 
In this case we have our following (x,n):
 
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>
+
<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:
 
Taking out our erroneous solutions:
−
</math>(30,10)<math>,</math>(50,6)<math>,</math>(172,2)<math>
+
<math>(30,10)</math>,<math>(50,6)</math>,<math>(172,2)</math>
 
Which also gives us three ways.
 
Which also gives us three ways.
−
Counting our cases: </math>3<math>+</math>3<math>=</math>6<math>,</math>D$
+
Counting our cases: <math>3</math>+<math>3</math>=<math>6</math>,<math>D</math>

Revision as of 12:14, 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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