Difference between revisions of "2012 AMC 10A Problems/Problem 24"
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{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #21]] and [[2012 AMC 10A Problems|2012 AMC 10A #24]]}} | {{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #21]] and [[2012 AMC 10A Problems|2012 AMC 10A #24]]}} | ||
Revision as of 20:43, 20 January 2017
- The following problem is from both the 2012 AMC 12A #21 and 2012 AMC 10A #24, so both problems redirect to this page.
Problem
Let ,
, and
be positive integers with
such that
and
.
What is ?
Solution
Add the two equations.
.
Now, this can be rearranged:
and factored:
,
, and
are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that
.
, since
is the biggest difference. It is impossible to determine by inspection whether
or
, or whether
or
.
We want to solve for , so take the two cases and solve them each for an expression in terms of
. Our two cases are
or
. Plug these values into one of the original equations to see if we can get an integer for
.
, after some algebra, simplifies to
.
is not divisible by
, so
is not an integer.
The other case gives , which simplifies to
. Thus,
and the answer is
.
See Also
2012 AMC 10A (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 |
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.