2011 AMC 12A Problems/Problem 20: Difference between revisions
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== Solution == | == Solution == | ||
From <math>f(1) = 0</math>, we know that <math>a+b+c = 0</math>. From the first inequality: | |||
<math>50 < 49a+7b+c < 60</math> | |||
<math>50 < 48a+6b < 60</math> | |||
<math>\frac{25}{3} < 8a+b < 10</math> | |||
Since <math>8a+b</math> must be an integer, it follows that <math>8a+b = 9</math>. Similarly, from the second inequality: | |||
<math>70 < 64a+8b+c < 80</math> | |||
<math>70 < 63a+7b < 80</math> | |||
<math>10 < 9a+b < \frac{80}{7}</math> | |||
And it follows that <math>9a+b = 11</math>. We now have a system of three equations. Solving it gives us <math>(a, b, c) = (2, -7, 5)</math>. From this, we find that | |||
<math>f(100) = 2(100)^2-7(100)+5 = 19295</math> | |||
And since <math>15000 < 19295 < 20000 \to 5000(3) < 19295 < 5000(4)</math>, we find that <math>k = \boxed{(\textbf{C}) 3}</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 13:07, 10 February 2011
Problem
Let 
, where 
, 
, and 
are integers. Suppose that 
, 
, 
, 
for some integer 
. What is ?
Solution
From 
, we know that 
. From the first inequality:
Since 
must be an integer, it follows that 
. Similarly, from the second inequality:
And it follows that 
. We now have a system of three equations. Solving it gives us 
. From this, we find that
And 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 | |
