Difference between revisions of "2011 AMC 12A Problems/Problem 20"
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Pretty esay, right? =) | Pretty esay, right? =) | ||
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{{AMC12 box|year=2011|num-b=19|num-a=21|ab=A}} | {{AMC12 box|year=2011|num-b=19|num-a=21|ab=A}} |
Revision as of 13:26, 20 February 2011
Problem
Let , where , , and are integers. Suppose that , , , for some integer . What is ?
Solution
From , we know that .
From the first inequality, we get . Subtracting from this gives us , and thus . Since must be an integer, it follows that .
Similarly, from the second inequality, we get . Again subtracting from this gives us , or . It follows from this that .
We now have a system of three equations: , , and . Solving gives us and from this we find that
Since , we find that .
See also
Pretty esay, right? =) SHUT UP
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |