Difference between revisions of "2005 AMC 10B Problems/Problem 24"
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== 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. | Note that all of the solutions used <math>a+b</math> or <math>a-b</math> as part of their solution. | ||
Revision as of 16:02, 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 
= $\boxed {(C)144}.
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
