2015 AMC 8 Problems/Problem 14

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

Solution 5

Since they want CONSECUTIVE odd numbers, it won't be hard to just list the sums out: $16=1+3+5+7$ $40=7+9+11+13$ $72=15+17+19+21$ $200=47+49+51+53$ All of the answer choices can be a sum of consecutive odd numbers except $100$ , so the answer is $\boxed{\textbf{(D)} 100}$ .

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Video Solution

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2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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