Difference between revisions of "2011 AMC 12A Problems/Problem 20"
(→See also) |
|||
Line 9: | Line 9: | ||
\textbf{(E)}\ 5 </math> | \textbf{(E)}\ 5 </math> | ||
− | == Solution == | + | == Solution 1 == |
From <math>f(1) = 0</math>, we know that <math>a+b+c = 0</math>. | From <math>f(1) = 0</math>, we know that <math>a+b+c = 0</math>. | ||
Line 18: | Line 18: | ||
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> | 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 < 19305 < 20000 \to 5000(3) < 19305 < 5000(4)</math>, we find that <math>k = 3 \rightarrow \boxed{ | + | Since <math>15000 < 19305 < 20000 \to 5000(3) < 19305 < 5000(4)</math>, we find that <math>k = 3 \rightarrow \boxed{\textbf{(C)}\ 3}</math>. |
+ | |||
+ | == Solution 2 == | ||
+ | <math>f(x)</math> is some non-monic quadratic with a root at <math>x=1</math>. Knowing this, we'll forget their silly <math>a</math>, <math>b</math>, and <math>c</math> and instead write it as <math>f(x)=p(x-1)(x-r)</math>. | ||
+ | |||
+ | <math>f(7)=6p(7-r)</math>, so <math>f(7)</math> is a multiple of 6. They say <math>f(7)</math> is between 50 and 60, exclusive. Notice that the only multiple of 6 in that range is 54. Thus, <math>f(7)=6p(7-r)=54</math>. | ||
+ | |||
+ | <math>f(8)=7p(8-r)</math>, so <math>f(8)</math> is a multiple of 7. They say <math>f(8)</math> is between 70 and 80, exclusive. Notice that the only multiple of 7 in that range is 77. Thus, <math>f(8)=7p(8-r)=77</math>. | ||
+ | |||
+ | Now, we solve a system of equations in two variables. | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 6p(7-r)&=54 \\ | ||
+ | 7p(8-r)&=77 \\ | ||
+ | \\ | ||
+ | p(7-r)&=9 \\ | ||
+ | p(8-r)&=11 \\ | ||
+ | \\ | ||
+ | 7p-pr&=9 \\ | ||
+ | 8p-pr&=11 \\ | ||
+ | \\ | ||
+ | (8p-pr)-(7p-pr)&=11-9 \\ | ||
+ | \\ | ||
+ | p&=2 \\ | ||
+ | \\ | ||
+ | 2(7-r)&=9 \\ | ||
+ | \\ | ||
+ | r&=2.5 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | <math>f(100)=2(100-1)(100-2.5)=19305 \implies k=3 \implies \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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:58, 26 July 2017
Contents
Problem
Let , where , , and are integers. Suppose that , , , for some integer . What is ?
Solution 1
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 .
Solution 2
is some non-monic quadratic with a root at . Knowing this, we'll forget their silly , , and and instead write it as .
, so is a multiple of 6. They say is between 50 and 60, exclusive. Notice that the only multiple of 6 in that range is 54. Thus, .
, so is a multiple of 7. They say is between 70 and 80, exclusive. Notice that the only multiple of 7 in that range is 77. Thus, .
Now, we solve a system of equations in two variables.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.