Difference between revisions of "2011 AMC 12A Problems/Problem 20"
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\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>. | ||
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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)}\ 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>. | |
− | {{AMC12 box|year=2011|num-b=19|num-a=21|ab=A}} | + | <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> | ||
+ | |||
+ | |||
+ | |||
+ | == Solution 3 (Essentially the same thing as Solution 1) == | ||
+ | |||
+ | So we know that <math>a,b,c</math> are integers so we can use this to our advantage | ||
+ | |||
+ | <math>\quad</math> | ||
+ | |||
+ | Using <math>f(1)=0</math>, we get the equation <math>a+b+c=0</math> and <math>f(7)=49a+7b+c=5X</math> where <math>X</math> is a decimal digit placeholder. (Ex. <math>X=2</math> provides the value <math>52</math>) | ||
+ | |||
+ | <math>\quad</math> | ||
+ | |||
+ | Solving for <math>b</math> using the system of equations, we get <math>48a+6b=5X</math> <math>\implies</math> <math>b=-8a+ \frac{5X}{6}</math> | ||
+ | |||
+ | <math>\quad</math> | ||
+ | |||
+ | Since we know that <math>a</math> and <math>b</math> are both integers, we know that <math>\frac{5X}{6}</math> <math>\in</math> <math>\mathbb{Z}</math> <math>\implies</math> <math>X=4</math> and by extension <math>b=-8a+9</math> | ||
+ | |||
+ | <math>\quad</math> | ||
+ | |||
+ | Attempting to solve for <math>b</math> again using the system <math>f(8)=64a+8b+c=7Y</math> (<math>Y</math> is another decimal digit placeholder),<math>f(1)=a+b+c=0</math> gives us <math>b=-9a+ \frac{7Y}{7}</math> <math>\implies</math> <math>Y=7</math> <math>\implies</math> <math>b=-9a+11</math> | ||
+ | |||
+ | <math>\quad</math> | ||
+ | |||
+ | This leads to <math>-8a+9=-9a+11</math> <math>\implies</math> <math>a=2</math> <math>\implies</math> <math>b=-7</math> | ||
+ | |||
+ | <math>\quad</math> | ||
+ | |||
+ | Plugging in the values of <math>a</math> and <math>b</math> into <math>f(1)=a+b+c=0</math>, we get <math>c=5</math> | ||
+ | |||
+ | <math>\quad</math> | ||
+ | |||
+ | Substituting the values of <math>a,b,c</math> into <math>f(100)=10000a+100b+c</math>, we get <math>f(100)=19305</math> and <math>5000k<19305<5000(k+1)</math> <math>\implies</math> <math>k=3</math> <math>\implies</math> | ||
+ | <math>\boxed{\textbf{(C)}\ 3}</math> | ||
+ | |||
+ | <math>\quad</math> | ||
+ | |||
+ | <math>\bf{Note}</math>: We can say that <math>f(7)=5X</math> and <math>f(8)=7Y</math> because we are given that <math>50<f(7)<60</math> and <math>70<f(8)<80</math> | ||
+ | |||
+ | |||
+ | == See also == {{AMC12 box|year=2011|num-b=19|num-a=21|ab=A}} |
Latest revision as of 16:36, 9 January 2024
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.
Solution 3 (Essentially the same thing as Solution 1)
So we know that are integers so we can use this to our advantage
Using , we get the equation and where is a decimal digit placeholder. (Ex. provides the value )
Solving for using the system of equations, we get
Since we know that and are both integers, we know that and by extension
Attempting to solve for again using the system ( is another decimal digit placeholder), gives us
This leads to
Plugging in the values of and into , we get
Substituting the values of into , we get and
: We can say that and because we are given that and
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 |