Difference between revisions of "2005 AMC 10B Problems/Problem 24"
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~Arcticturn | ~Arcticturn | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/ybsryPSpGfM | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == |
Revision as of 08:57, 24 March 2023
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
Video Solution
~savannahsolver
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.