2005 AMC 10B Problems/Problem 24
Contents
[hide]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
. Since
, substituting and factoring, we get
. Therefore,
and
must be an integer. Therefore 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
since
and
are positive. Finally we come to the conclusion that
and
so
=
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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