1987 AIME Problems/Problem 12
Problem
Let be the smallest integer whose cube root is of the form
, where
is a positive integer and
is a positive real number less than
. Find
.
Solution
In order to keep as small as possible, we need to make
as small as possible.
. Since
and
is an integer, we must have that
. This means that the smallest possible
should be quite a bit smaller than 1000, so in particular
should be less than 1, so
and
.
, so we must have
. Since we want to minimize
, we take
. Then for any positive value of
,
, so it possible for
to be less than
. However, we still have to make sure a sufficiently small
exists.
In light of the equation , we need to choose
as small as possible to insure a small enough
. The smallest possible value for
is 1, when
. Then for this value of
,
, and we're set. The answer is
.
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |