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

(Solution)
(Solution)
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<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>(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 13:13, 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: $3$+$3$=$6$,$D$