Difference between revisions of "2015 AMC 8 Problems/Problem 14"
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===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>100</math>, so the answer is <math>\boxed{\textbf{(D)}~100}</math>. |
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==See Also== | ==See Also== | ||
{{AMC8 box|year=2015|num-b=13|num-a=15}} | {{AMC8 box|year=2015|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:31, 25 November 2015
Which of the following integers cannot be written as the sum of four consecutive odd integers?
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 
, so the answer is 
.
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. 
