Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 12: Line 12:
 
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>,</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</math>
+
Counting our cases: </math>3<math>+</math>3<math>=</math>6<math>,</math>D$

Revision as of 13:14, 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$ (Error compiling LaTeX. Unknown error_msg)x$and there are$n$numbers, this is the following sum:$x+x+1+x+2+...+x+(n-1)=$$ (Error compiling LaTeX. Unknown error_msg)xn+1+2+3+...+n-1=$$ (Error compiling LaTeX. Unknown error_msg)xn+n(n-1)/2=$$ (Error compiling LaTeX. Unknown error_msg)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:$3$+$3$=$6$,$D$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_18&oldid=76694"