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== | ||
===Solution 1=== | ===Solution 1=== | ||
| − | 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>. | + | 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=== | ||
| − | 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>100</math>, so the answer is <math>\boxed{\textbf{(D)}~100}</math>. | + | 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=== | ||
| + | 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=== | ||
| + | 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=== | ||
| + | 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
Contents
Problem
Which of the following integers cannot be written as the sum of four consecutive odd integers?
Solutions
Solution 1
Let our 
numbers be 
, where 
is odd. Then, our sum is 
. The only answer choice that cannot be written as , where is odd, is 
.
Solution 2
If the four consecutive odd integers are 
and 
; then, the sum is 
. All the integers are divisible by 
except 
.
Solution 3
If the four consecutive odd integers are 
and 
, the sum is 
, and divided by gives 
. This means that must be even. The only integer that does not give an even integer when divided by is 
, so the answer is .
Solution 4
From Solution 1, we have the sum of the numbers to be equal to 
. Taking mod 8 gives us 
for some residue 
and for some odd integer . Since 
, we can express it as the equation 
for some integer 
. Multiplying 4 to each side of the equation yields 
, and taking mod 8 gets us 
, so 
. All the answer choices except choice D is a multiple of 8, and since 100 satisfies all the conditions the answer is .
Solution 5
Since they want CONSECUTIVE odd numbers, it won't be hard to just list the sums out:
![\[16=1+3+5+7\]](http://latex.artofproblemsolving.com/6/2/0/6204f662662450cdbe75b27565b79cd88ee0dd55.png)
![\[40=7+9+11+13\]](http://latex.artofproblemsolving.com/3/1/2/312d83a548dde49ddd0a0c086d73de8fab761aee.png)
![\[72=15+17+19+21\]](http://latex.artofproblemsolving.com/5/7/c/57c906e973e0620a3e0d75e2613952eb406c8791.png)
![\[200=47+49+51+53\]](http://latex.artofproblemsolving.com/9/f/4/9f435a1f68f1b821fae6493e28b37c8e5996b807.png)
All of the answer choices can be a sum of consecutive odd numbers except , so the answer is 
.
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
