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

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==Problem==
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Which of the following integers cannot be written as the sum of four consecutive odd integers?
 
Which of the following integers cannot be written as the sum of four consecutive odd integers?
  
 
<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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==Solutions==
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===Solution 1===
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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>.
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===Solution 2===
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If the four consecutive odd integers are <math>2n-3,~ 2n-1, ~2n+1</math> and <math>2n+3</math> then the sum is <math>8n</math>. All the integers are divisible by <math>8</math> except <math>\boxed{\textbf{(D)}~100}</math>.
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===Solution 3===
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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>.
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===Solution 4===
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From Solution 1, we have the sum of the <math>4</math> numbers to be equal to <math>4n + 12</math>. Taking mod 8 gives us <math>4n + 4 \equiv b \pmod8</math> for some residue <math>b</math> and for some odd integer <math>n</math>. Since <math>n \equiv 1 \pmod{2}</math>, we can express it as the equation <math>n = 2a + 1</math> for some integer <math>a</math>. Multiplying 4 to each side of the equation yields <math>4n = 8a + 4</math>, and taking mod 8 gets us <math>4n \equiv 4 \pmod{8}</math>, so <math>b = 0</math>. All the answer choices except choice D is a multiple of 8, and since 100 satisfies all the conditions the answer is <math>\boxed{\textbf{(D)}~100}</math>.
  
 
==See Also==
 
==See Also==
  
{{AMC8 box|year=2015|num-b=13|after=15}}
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{{AMC8 box|year=2015|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:40, 16 January 2021

Problem

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

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

Solutions

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

Solution 4

From Solution 1, we have the sum of the $4$ numbers to be equal to $4n + 12$. Taking mod 8 gives us $4n + 4 \equiv b \pmod8$ for some residue $b$ and for some odd integer $n$. Since $n \equiv 1 \pmod{2}$, we can express it as the equation $n = 2a + 1$ for some integer $a$. Multiplying 4 to each side of the equation yields $4n = 8a + 4$, and taking mod 8 gets us $4n \equiv 4 \pmod{8}$, so $b = 0$. All the answer choices except choice D is a multiple of 8, and since 100 satisfies all the conditions 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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