Difference between revisions of "2015 AMC 8 Problems/Problem 14"
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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>. | + | 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=== |
Revision as of 08:18, 26 March 2023
Contents
[hide]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: All of the answer choices can be a sum of consecutive odd numbers except , so the answer is
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 |
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