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Difference between revisions of "2015 AMC 8 Problems/Problem 14"
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===Solution 1===
 
===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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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>.
  
 
===Solution 2===
 
===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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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>.
  
 
===Solution 3===
 
===Solution 3===
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===Solution 4===
 
===Solution 4===
 
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>.
 
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>.
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 +
===Solution 5===
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Since they want CONSECUTIVE odd numbers, it won't be hard to just list the sums out:
 +
<cmath>16=1+3+5+7</cmath>
 +
<cmath>40=7+9+11+13</cmath>
 +
<cmath>72=15+17+19+21</cmath>
 +
<cmath>200=47+49+51+53</cmath>
 +
All of the answer choices can be a sum of consecutive odd numbers except <math>100</math>, so the answer is <math>\boxed{\textbf{(D)} 100}</math>.
 +
 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/Xc7EYnJL5_c
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution==
 +
https://youtu.be/khCUYxEJFQg
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 15:46, 14 May 2023

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

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/Xc7EYnJL5_c

~Education, the Study of Everything

Video Solution

https://youtu.be/khCUYxEJFQg

~savannahsolver

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
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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