Difference between revisions of "2005 AMC 10B Problems/Problem 24"
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What is <math>x + y + m</math>? | What is <math>x + y + m</math>? | ||
− | <math>\ | + | <math>\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154 </math> |
− | == Solution == | + | == Solution 1 == |
− | Let <math>x = 10a+b, y = 10b+a</math>. The given conditions imply <math>x>y</math>, which implies <math>a>b</math>, and they also imply that both <math>a</math> and <math>b</math> are nonzero. Then <math>x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2</math>. Since this must be a perfect square, all the exponents in its prime factorization must be even. <math>99</math> factorizes into <math>3^2 \cdot 11</math>, so <math>11|(a-b)(a+b)</math>. However, the maximum value of <math>a-b</math> is <math>9-1=8</math>, so <math>11|a+b</math>. The maximum of <math>a+b</math> is <math>9+8=17</math>, so <math>a+b=11</math>. Then we have <math>33^2(a-b) = m^2</math>, so <math>a-b</math> is a perfect square, but the only perfect squares that are within our bound on <math>a-b</math> are <math>1</math> and <math>4</math>. We know <math>a+b=11</math>, and, for <math>a-b=1</math>, adding equations to eliminate <math>b</math> gives us <math>2a=12 \Longrightarrow a=6, b=5</math>. Testing <math>a-b=4</math> gives us <math>2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{ | + | Let <math>x = 10a+b, y = 10b+a</math>. The given conditions imply <math>x>y</math>, which implies <math>a>b</math>, and they also imply that both <math>a</math> and <math>b</math> are nonzero. |
+ | |||
+ | Then, <math>x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2</math>. | ||
+ | |||
+ | Since this must be a perfect square, all the exponents in its prime factorization must be even. <math>99</math> factorizes into <math>3^2 \cdot 11</math>, so <math>11|(a-b)(a+b)</math>. However, the maximum value of <math>a-b</math> is <math>9-1=8</math>, so <math>11|a+b</math>. The maximum value of <math>a+b</math> is <math>9+8=17</math>, so <math>a+b=11</math>. | ||
+ | |||
+ | Then, we have <math>33^2(a-b) = m^2</math>, so <math>a-b</math> is a perfect square, but the only perfect squares that are within our bound on <math>a-b</math> are <math>1</math> and <math>4</math>. We know <math>a+b=11</math>, and, for <math>a-b=1</math>, adding equations to eliminate <math>b</math> gives us <math>2a=12 \Longrightarrow a=6, b=5</math>. Testing <math>a-b=4</math> gives us <math>2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{7}{2}</math>, which is impossible, as <math>a</math> and <math>b</math> must be digits. Therefore, <math>(a,b) = (6,5)</math>, and <math>x+y+m=\boxed{\textbf{(E) }154}</math>. | ||
== Solution 2 == | == Solution 2 == | ||
− | The first steps are the same as | + | The first steps are the same as Solution 1. Let <math>x = 10a+b, y = 10b+a</math>, where we know that a and b are digits (whole numbers less than <math>10</math>). |
+ | |||
+ | Like Solution 1, we end up getting <math>(9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2</math>. This is where the solution diverges. | ||
− | We know that the left side of the equation is a perfect square because m is an integer. If we factor 99 into its prime factors, we get <math>3^2\cdot 11</math>. In order to get a perfect square on the left side, <math>(a-b)(a+b)</math> must make both prime exponents even. Because the a and b are digits, a simple guess would be that <math>(a+b)</math> (the bigger number) equals 11 while <math>(a-b)</math> is a factor of nine (1 or 9). The correct guesses are <math>a = 6, b = 5</math> causing <math>x = 65, y = 56,</math> and <math>m = 33</math>. The sum of the numbers is <math>\boxed{\textbf{(E) }154}</math> | + | We know that the left side of the equation is a perfect square because <math>m</math> is an integer. If we factor <math>99</math> into its prime factors, we get <math>3^2\cdot 11</math>. In order to get a perfect square on the left side, <math>(a-b)(a+b)</math> must make both prime exponents even. Because the a and b are digits, a simple guess would be that <math>(a+b)</math> (the bigger number) equals <math>11</math> while <math>(a-b)</math> is a factor of nine (1 or 9). The correct guesses are <math>a = 6, b = 5</math> causing <math>x = 65, y = 56,</math> and <math>m = 33</math>. The sum of the numbers is <math>\boxed{\textbf{(E) }154}</math> |
== Solution 3 == | == Solution 3 == | ||
− | Once again, the solution is quite similar as the above solutions. Since <math>x</math> and <math>y</math> are two digit integers, we can write <math>x = 10a+b, y = 10b+a</math> and because <math>x^2 - y^2 = (x-y)(x+y)</math>, substituting and factoring, we get <math>99(a+b)(a-b) = m^2</math>. Therefore, <math>(a+b)(a-b) = \frac{m^2}{99}</math> and <math>\frac{m^2}{99}</math> must be an integer. A quick strategy is to find the smallest such integer <math>m</math> such that <math>\frac{m^2}{99}</math> is an integer. We notice that 99 has a prime factorization of <math>3^2 \cdot 11.</math> Let <math>m^2 = n.</math> Since we need a perfect square and 3 is already squared, we just need to square 11. So <math>3^2 \cdot 11^2</math> gives us 1089 as <math>n</math> and <math>m = \sqrt{1089} = 33.</math> We now get the equation <math>(x-y)(x+y) = 33^2</math>, which we can also write as <math>(x-y)(x+y) = 11^2 \cdot 3^2</math>. A very simple guess assumes that <math>x-y=3^2</math> and <math>x+y=11^2</math> since <math>x</math> and <math>y</math> are positive. Finally, we come to the conclusion that <math>x=65</math> and <math>y=56</math>, so <math>x+y+m</math> <math>=</math> <math>\boxed{\textbf{(E) }154}</math>. | + | Once again, the solution is quite similar as the above solutions. Since <math>x</math> and <math>y</math> are two digit integers, we can write <math>x = 10a+b, y = 10b+a</math> and because <math>x^2 - y^2 = (x-y)(x+y)</math>, substituting and factoring, we get <math>99(a+b)(a-b) = m^2</math>. |
+ | |||
+ | Therefore, <math>(a+b)(a-b) = \frac{m^2}{99}</math> and <math>\frac{m^2}{99}</math> must be an integer. A quick strategy is to find the smallest such integer <math>m</math> such that <math>\frac{m^2}{99}</math> is an integer. We notice that 99 has a prime factorization of <math>3^2 \cdot 11.</math> | ||
+ | |||
+ | Let <math>m^2 = n.</math> Since we need a perfect square and 3 is already squared, we just need to square 11. So <math>3^2 \cdot 11^2</math> gives us 1089 as <math>n</math> and <math>m = \sqrt{1089} = 33.</math> We now get the equation <math>(x-y)(x+y) = 33^2</math>, which we can also write as <math>(x-y)(x+y) = 11^2 \cdot 3^2</math>. | ||
+ | |||
+ | A very simple guess assumes that <math>x-y=3^2</math> and <math>x+y=11^2</math> since <math>x</math> and <math>y</math> are positive. Finally, we come to the conclusion that <math>x=65</math> and <math>y=56</math>, so <math>x+y+m</math> <math>=</math> <math>\boxed{\textbf{(E) }154}</math>. | ||
+ | |||
+ | Note that all of the solutions used <math>a+b</math> or <math>a-b</math> as part of their solution. | ||
+ | |||
+ | == Solution 4 == | ||
+ | Continue the same as Solution <math>3</math> until we get <math>33</math>. Knowing that <math>33^2 = 1089</math>, we have narrowed down our Pythagorean triples. We know that the <math>2</math> other squares should be larger than <math>33^2</math>, so we can start testing. | ||
+ | |||
+ | If we start testing the <math>40</math>s, it is fruitless since the closest to <math>33^2</math> would be <math>33 - 45 - 54</math> which is not a Pythagorean triple. We can start by testing out the <math>50</math>s, and it turns our that <math>33 - 56 - 65</math> is a Pythagorean triple. Therefore, our answer is <math>33+56+65</math> = <math>\boxed{\textbf{(C) }144}</math>. | ||
+ | |||
+ | ~Arcticturn | ||
== See Also == | == See Also == |
Latest revision as of 17:04, 16 December 2021
Problem
Let and be two-digit integers such that is obtained by reversing the digits of . The integers and satisfy for some positive integer . What is ?
Solution 1
Let . The given conditions imply , which implies , and they also imply that both and are nonzero.
Then, .
Since this must be a perfect square, all the exponents in its prime factorization must be even. factorizes into , so . However, the maximum value of is , so . The maximum value of is , so .
Then, we have , so is a perfect square, but the only perfect squares that are within our bound on are and . We know , and, for , adding equations to eliminate gives us . Testing gives us , which is impossible, as and must be digits. Therefore, , and .
Solution 2
The first steps are the same as Solution 1. Let , where we know that a and b are digits (whole numbers less than ).
Like Solution 1, we end up getting . This is where the solution diverges.
We know that the left side of the equation is a perfect square because is an integer. If we factor into its prime factors, we get . In order to get a perfect square on the left side, must make both prime exponents even. Because the a and b are digits, a simple guess would be that (the bigger number) equals while is a factor of nine (1 or 9). The correct guesses are causing and . The sum of the numbers is
Solution 3
Once again, the solution is quite similar as the above solutions. Since and are two digit integers, we can write and because , substituting and factoring, we get .
Therefore, and must be an integer. A quick strategy is to find the smallest such integer such that is an integer. We notice that 99 has a prime factorization of
Let Since we need a perfect square and 3 is already squared, we just need to square 11. So gives us 1089 as and We now get the equation , which we can also write as .
A very simple guess assumes that and since and are positive. Finally, we come to the conclusion that and , so .
Note that all of the solutions used or as part of their solution.
Solution 4
Continue the same as Solution until we get . Knowing that , we have narrowed down our Pythagorean triples. We know that the other squares should be larger than , so we can start testing.
If we start testing the s, it is fruitless since the closest to would be which is not a Pythagorean triple. We can start by testing out the s, and it turns our that is a Pythagorean triple. Therefore, our answer is = .
~Arcticturn
See Also
2005 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 10 Problems and Solutions |
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