Difference between revisions of "1998 AIME Problems/Problem 14"
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Then since <math>\dfrac{m+n+2}{(m+2)(n+2)}>\dfrac{1}{2}</math> for <math>m=1,2,</math> we fix <math>m=3.</math> | Then since <math>\dfrac{m+n+2}{(m+2)(n+2)}>\dfrac{1}{2}</math> for <math>m=1,2,</math> we fix <math>m=3.</math> | ||
− | <cmath>\implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(n+ | + | <cmath>\implies \dfrac{1}{p}=\dfrac{1}{2}-\dfrac{2(n+5)}{5(n+2)}=\dfrac{n-10}{10(n+2)},</cmath> |
where we simply let <math>n=11</math> to achieve <math>p=\boxed{130}.</math> | where we simply let <math>n=11</math> to achieve <math>p=\boxed{130}.</math> | ||
~Generic_Username | ~Generic_Username | ||
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== See also == | == See also == |
Latest revision as of 21:05, 29 May 2023
Contents
Problem
An rectangular box has half the volume of an
rectangular box, where
and
are integers, and
What is the largest possible value of
?
Solution 1
Let’s solve for :
Clearly, we want to minimize the denominator, so we test . The possible pairs of factors of
are
. These give
and
respectively. Substituting into the numerator, we see that the first pair gives
, while the second pair gives
. We now check that
is optimal, setting
,
in order to simplify calculations. Since
We have
Where we see
gives us our maximum value of
.
- Note that
assumes
, but this is clear as
and similarly for
.
Solution 2
Similarly as above, we solve for but we express the denominator differently:
Hence, it suffices to maximize
under the conditions that
is a positive integer.
Then since for
we fix
where we simply let
to achieve
~Generic_Username
See also
1998 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.