Difference between revisions of "2011 AMC 12B Problems/Problem 21"
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<math>\sqrt{xy} = 10 b+a</math> | <math>\sqrt{xy} = 10 b+a</math> | ||
− | Squaring the first and second equations, <math> | + | Squaring the first and second equations, <math>\frac{x^2 + 2xy + y^2}{4}=100 a^2 + 20 ab + b^2</math> |
<math>xy = 100b^2 + 20ab + a^2</math> | <math>xy = 100b^2 + 20ab + a^2</math> | ||
− | <math>\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)</math> | + | Subtracting the previous two equations, <math>\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)</math> |
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Note that for x-y to be an integer, <math>(a^2 - b^2)</math> has to be <math>11n</math> for some perfect square <math>n</math>. Since <math>a</math> is at most <math>9</math>, <math>n = 1</math> or <math>4</math> | Note that for x-y to be an integer, <math>(a^2 - b^2)</math> has to be <math>11n</math> for some perfect square <math>n</math>. Since <math>a</math> is at most <math>9</math>, <math>n = 1</math> or <math>4</math> | ||
− | If <math>n = 1</math>, <math>|x-y| = 66</math>, if <math>n = 4</math>, <math>|x-y| = 132</math>. In AMC, we are done. Otherwise, we need to show that <math>a^2 -b^2 = 44</math> is impossible. | + | If <math>n = 1</math>, <math>|x-y| = 66</math>, if <math>n = 4</math>, <math>|x-y| = 132</math>. In AMC, we are done. Otherwise, we need to show that <math>n=4</math>, or <math>a^2 -b^2 = 44</math> is impossible. |
<math>(a-b)(a+b) = 44</math> -> <math>a-b = 1</math>, or <math>2</math> or <math>4</math> and <math>a+b = 44</math>, <math>22</math>, <math>11</math> respectively. And since <math>a+b \le 18</math>, <math>a+b = 11</math>, <math>a-b = 4</math>, but there is no integer solution for <math>a</math>, <math>b</math>. | <math>(a-b)(a+b) = 44</math> -> <math>a-b = 1</math>, or <math>2</math> or <math>4</math> and <math>a+b = 44</math>, <math>22</math>, <math>11</math> respectively. And since <math>a+b \le 18</math>, <math>a+b = 11</math>, <math>a-b = 4</math>, but there is no integer solution for <math>a</math>, <math>b</math>. | ||
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Let <math>(x+y)/2 = 10a + b</math> and <math>\sqrt{xy} = 10b + a</math>. By AM-GM we know that <math>a \ge b</math>. Squaring and multiplying by 4 on the first equation we get <math>x^2 + y^2 + 2xy = 400a^2 + 4b^2 + 80ab</math>. Squaring and multiplying the second equation by 4 we get <math>4xy = 400b^2 + 4a^2 + 80ab</math>. Subtracting we get <math>(x-y)^2 = 396(a^2 - b^2)</math>. Note that <math>396 = 2^2 \cdot 3^2 \cdot 11</math>. So to make it a perfect square <math>a^2 - b^2 = 11</math>. From [[difference of squares]], we see that <math>a = 6</math> and <math>b = 5</math>. So the answer is <math>3 \cdot 2 \cdot 11 = 66</math>. | Let <math>(x+y)/2 = 10a + b</math> and <math>\sqrt{xy} = 10b + a</math>. By AM-GM we know that <math>a \ge b</math>. Squaring and multiplying by 4 on the first equation we get <math>x^2 + y^2 + 2xy = 400a^2 + 4b^2 + 80ab</math>. Squaring and multiplying the second equation by 4 we get <math>4xy = 400b^2 + 4a^2 + 80ab</math>. Subtracting we get <math>(x-y)^2 = 396(a^2 - b^2)</math>. Note that <math>396 = 2^2 \cdot 3^2 \cdot 11</math>. So to make it a perfect square <math>a^2 - b^2 = 11</math>. From [[difference of squares]], we see that <math>a = 6</math> and <math>b = 5</math>. So the answer is <math>3 \cdot 2 \cdot 11 = 66</math>. | ||
~coolmath_2018 | ~coolmath_2018 | ||
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+ | ==Solution 3 (whee!) == | ||
+ | Let the 2 digit number be <math>10a+b</math>. We know <math>x+y=20a+2b</math> and <math>\sqrt{xy} = 10b+a</math>. | ||
+ | |||
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+ | Thus. <math>x^2+2xy+y^2 = 400a^2+80ab+4b^2</math> and <math>xy = 100b^2+20ab+a^2</math>. Notice that <math>|x-y| = \sqrt{(x-y)^2}</math>. We know that <math>(x-y)^2 = (x+y)^2-4xy = 396a^2-396b^2</math>. Then, <math>|x-y| = 6\sqrt{11}\cdot \sqrt{b^2-a^2}</math>. Clearly, <math>|x-y|</math> must be a multiple of 11, so the only possible answer is <math>\boxed{\textbf{(D) }66}</math>. | ||
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+ | -skibbysiggy | ||
== See also == | == See also == |
Latest revision as of 15:17, 5 September 2024
Contents
Problem
The arithmetic mean of two distinct positive integers and is a two-digit integer. The geometric mean of and is obtained by reversing the digits of the arithmetic mean. What is ?
Solution 1
Answer: (D)
for some , .
Squaring the first and second equations,
Subtracting the previous two equations,
Note that for x-y to be an integer, has to be for some perfect square . Since is at most , or
If , , if , . In AMC, we are done. Otherwise, we need to show that , or is impossible.
-> , or or and , , respectively. And since , , , but there is no integer solution for , .
Short Cut
We can arrive at using the method above. Because we know that is an integer, it must be a multiple of 6 and 11. Hence the answer is
In addition: Note that with may be obtained with and as .
Sidenote
It is easy to see that is the only solution. This yields . Their arithmetic mean is and their geometric mean is .
Solution 2
Let and . By AM-GM we know that . Squaring and multiplying by 4 on the first equation we get . Squaring and multiplying the second equation by 4 we get . Subtracting we get . Note that . So to make it a perfect square . From difference of squares, we see that and . So the answer is . ~coolmath_2018
Solution 3 (whee!)
Let the 2 digit number be . We know and .
Thus. and . Notice that . We know that . Then, . Clearly, must be a multiple of 11, so the only possible answer is .
-skibbysiggy
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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 |
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