Difference between revisions of "2012 AMC 10B Problems/Problem 9"
m (→Problem 9) |
(→Solution) |
||
Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
− | Out of the first two integers, it's possible for both to be | + | Out of the first two integers, it's possible for both to be even: for example, <math>10 + 16 = 26.</math> But the next two integers, when added, increase the sum by <math>15,</math> which is odd, so one of them must be odd and the other must be even: for example, <math>3 + 12 = 15.</math> Finally, the next two integers increase the sum by <math>16,</math> which is even, so we can have both be even: for example, <math>2 + 14 = 16.</math> Therefore, <math>\boxed{\textbf{(A) } 1}</math> is the minimum number of integers that must be odd. |
− | |||
− | |||
==See Also== | ==See Also== |
Revision as of 19:27, 20 September 2017
Problem 9
Two integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of the previous four integers the sum is 57. What is the minimum number of odd integers among the 6 integers?
Solution
Out of the first two integers, it's possible for both to be even: for example, But the next two integers, when added, increase the sum by which is odd, so one of them must be odd and the other must be even: for example, Finally, the next two integers increase the sum by which is even, so we can have both be even: for example, Therefore, is the minimum number of integers that must be odd.
See Also
2012 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.