Difference between revisions of "2005 AMC 10B Problems/Problem 24"
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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 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> | ||
+ | |||
+ | == 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> | ||
== See Also == | == See Also == |
Revision as of 18:01, 9 June 2016
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
Let , without loss of generality with
. Then
. It follows that
, but
so
. Then we have
. Thus
is a perfect square. Also, because
and
have the same parity,
is a one-digit odd perfect square, namely
or
. The latter case gives
, which does not work. The former case gives
, which works, and we have
.
Solution 2
The first steps are the same as above. Let , where we know that a and b are digits (whole numbers less than 10). Like above, we end up getting
. 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 . 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 11 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
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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