Difference between revisions of "2011 AMC 12A Problems/Problem 20"
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Similarly, from the second inequality, we get <math>70 < 64a+8b+c < 80</math>. Again subtracting <math>a+b+c = 0</math> from this gives us <math>70 < 63a+7b < 80</math>, or <math>10 < 9a+b < \frac{80}{7}</math>. It follows from this that <math>9a+b = 11</math>. | Similarly, from the second inequality, we get <math>70 < 64a+8b+c < 80</math>. Again subtracting <math>a+b+c = 0</math> from this gives us <math>70 < 63a+7b < 80</math>, or <math>10 < 9a+b < \frac{80}{7}</math>. It follows from this that <math>9a+b = 11</math>. | ||
− | We now have a system of three equations: <math>a+b+c = 0</math>, <math>8a+b = 9</math>, and <math>9a+b = 11</math>. Solving gives us <math>(a, b, c) = (2, -7, 5)</math> and from this we find that <math>f(100) = 2(100)^2-7(100)+5 = | + | We now have a system of three equations: <math>a+b+c = 0</math>, <math>8a+b = 9</math>, and <math>9a+b = 11</math>. Solving gives us <math>(a, b, c) = (2, -7, 5)</math> and from this we find that <math>f(100) = 2(100)^2-7(100)+5 = 19305</math> |
− | Since <math>15000 < | + | Since <math>15000 < 19305 < 20000 \to 5000(3) < 19305 < 5000(4)</math>, we find that <math>k = 3 \rightarrow \boxed{(\textbf{C})}</math>. |
== See also == | == See also == | ||
{{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 12:49, 24 June 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
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 |