Difference between revisions of "2015 AMC 8 Problems/Problem 14"
Line 3: | Line 3: | ||
<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> | ||
− | ===Solution=== | + | ===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=== | ||
+ | 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>. | ||
==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 17:31, 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.