Difference between revisions of "2015 AMC 8 Problems/Problem 14"

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<math>\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}</math>
 
<math>\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}</math>
  
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===Solution===
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Let our <math>4</math> numbers be <math>n, n+2, n+4, n+6</math>, where <math>n</math> is odd. Then our sum is <math>4n+12</math>. The only answer choice that cannot be written as <math>4n+12</math>, where <math>n</math> is odd, is <math>\boxed{\textbf{(D)}\text{ 100}}</math>.
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=2015|num-b=13|num-a=15}}
 
{{AMC8 box|year=2015|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:27, 25 November 2015

Which of the following integers cannot be written as the sum of four consecutive odd integers?

$\textbf{(A)}\text{ 16}\quad\textbf{(B)}\text{ 40}\quad\textbf{(C)}\text{ 72}\quad\textbf{(D)}\text{ 100}\quad\textbf{(E)}\text{ 200}$

Solution

Let our $4$ numbers be $n, n+2, n+4, n+6$, where $n$ is odd. Then our sum is $4n+12$. The only answer choice that cannot be written as $4n+12$, where $n$ is odd, is $\boxed{\textbf{(D)}\text{ 100}}$.

See Also

2015 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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 AJHSME/AMC 8 Problems and Solutions

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