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Difference between revisions of "2016 AMC 12B Problems/Problem 16"

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(→‎Solution 2)
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<math>2\cdot 3\cdot 5\cdot 23=(a-b+1)(a+b)</math>
 
<math>2\cdot 3\cdot 5\cdot 23=(a-b+1)(a+b)</math>
  
−
If we factor <math>345</math> into all of its factor groups <math>(\text{exg}~ (5,69) ~\text{or} ~(15,23))</math> we will have several ordered pairs <math>(c,d)</math> where <math>c<d</math>.
+
Let <math>c=a-b+1</math> and <math>d=a+b</math>  
  
−
The number of possible values for <math>c</math> is half the number of factors of <math>345</math> which is <math>\frac{1}{2}\cdot2\cdot2\cdot2\cdot2=8</math>
+
<math>2\cdot 3\cdot 5\cdot 23=c\cdot d\)
  
−
However, we have one extraneous case of <math>(1,345)</math> because here, <math>a=b</math> and we have the sum of one consecutive number which is not allowed by the question.
+
If we factor </math>690<math> into all of its factor groups </math>(\text{exg}~ (10,69) ~\text{or} ~(15,46))<math> we will have several ordered pairs </math>(c,d)<math> where </math>c<d<math>
  
−
Thus the answer is <math>8-1=7</math>
+
The number of possible values for </math>c<math> is half the number of factors of </math>345<math> which is </math>\frac{1}{2}\cdot2\cdot2\cdot2\cdot2=8<math>  
  
−
<math>\boxed{\textbf{(E)} \,7}</math>.
+
However, we have one extraneous case of </math>(1,690)<math> because here, </math>a=b<math> and we have the sum of one consecutive number which is not allowed by the question.
 +
 
 +
Thus the answer is </math>8-1=7<math>
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 +
</math>\boxed{\textbf{(E)} \,7}$.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=15|num-a=17}}
 
{{AMC12 box|year=2016|ab=B|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:03, 22 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

We proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number of consecutive numbers.

For the first case, we can cleverly choose the convenient form of our sequence to be \[a-n,\cdots, a-1, a, a+1, \cdots, a+n\]

because then our sum will just be $(2n+1)a$. We now have \[(2n+1)a = 345\] and $a$ will have a solution when $\frac{345}{2n+1}$ is an integer, namely when $2n+1$ is a divisor of 345. We check that \[2n+1 = 3, 5, 15, 23\] work, and no more, because $2n+1=1$ does not satisfy the requirements of two or more consecutive integers, and when $2n+1$ equals the next biggest factor, $69$, there must be negative integers in the sequence. Our solutions are $\{114,115, 116\}, \{67, \cdots, 71\}, \{16, \cdots, 30\}, \{4, \cdots, 26\}$.

For the even cases, we choose our sequence to be of the form: \[a-(n-1), \cdots, a, a+1, \cdots, a+n\] so the sum is $\frac{(2n)(2a+1)}{2} = n(2a+1)$. In this case, we find our solutions to be $\{172, 173\}, \{55,\cdots, 60\}, \{30,\cdots, 39\}$.

We have found all 7 solutions and our answer is $\boxed{\textbf{(E)} \, 7}$.

Solution 2

The sum from $a$ to $b$ where $a$ and $b$ are integers and $a>b$ is

$S=\dfrac{(a-b+1)(a+b)}{2}$

$345=\dfrac{(a-b+1)(a+b)}{2}$

$2\cdot 3\cdot 5\cdot 23=(a-b+1)(a+b)$

Let $c=a-b+1$ and $d=a+b$

$2\cdot 3\cdot 5\cdot 23=c\cdot d\)

If we factor$ (Error compiling LaTeX. Unknown error_msg)690$into all of its factor groups$(\text{exg}~ (10,69) ~\text{or} ~(15,46))$we will have several ordered pairs$(c,d)$where$c<d$The number of possible values for$c$is half the number of factors of$345$which is$\frac{1}{2}\cdot2\cdot2\cdot2\cdot2=8$However, we have one extraneous case of$(1,690)$because here,$a=b$and we have the sum of one consecutive number which is not allowed by the question.

Thus the answer is$ (Error compiling LaTeX. Unknown error_msg)8-1=7$$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(E)} \,7}$.

See Also

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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