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

(Solution 2)
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===Solution 3===
 
===Solution 3===
If the four consecutive odd integers are <math>a,~ a+2, ~a+4</math> and <math>a+6</math> and the sum is <math>b</math> then <math>b</math> divided by <math>4</math> is equal to <math>a+3</math>. This means that <math>a+3</math> must be even. The only integer that does not give an even integer when divided by <math>4</math> is <math>100</math>, so the answer is <math>\boxed{\textbf{(D)}~100}</math>.
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If the four consecutive odd integers are <math>a,~ a+2, ~a+4</math> and <math>a+6</math>, the sum is <math>4a+12</math>, and <math>4a+12</math> divided by <math>4</math> gives <math>a+3</math>. This means that <math>a+3</math> must be even. The only integer that does not give an even integer when divided by <math>4</math> is <math>100</math>, so the answer is <math>\boxed{\textbf{(D)}~100}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 13:08, 30 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 1

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 $\boxed{\textbf{(D)}\text{ 100}}$.

Solution 2

If the four consecutive odd integers are $2n-3,~ 2n-1, ~2n+1$ and $2n+3$ then the sum is $8n$. All the integers are divisible by $8$ except $\boxed{\textbf{(D)}~100}$.

Solution 3

If the four consecutive odd integers are $a,~ a+2, ~a+4$ and $a+6$, the sum is $4a+12$, and $4a+12$ divided by $4$ gives $a+3$. This means that $a+3$ must be even. The only integer that does not give an even integer when divided by $4$ is $100$, so the answer is $\boxed{\textbf{(D)}~100}$.

See Also

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

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