Difference between revisions of "2015 AMC 8 Problems/Problem 14"
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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 = | + | 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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==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 12:53, 3 July 2019
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 
.
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 .
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. 
