Difference between revisions of "2015 AMC 8 Problems/Problem 14"
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<cmath>200=47+49+51+53</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>. | 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>. | ||
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+ | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
+ | https://youtu.be/Xc7EYnJL5_c | ||
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+ | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== |
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: 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.