1987 AIME Problems/Problem 12
Contents
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 1
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. 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 is 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 ensure a small enough
. The smallest possible value for
is 1, when
. Then for this value of
,
, and we're set. The answer is
.
Solution 2
To minimize , we should minimize
. We have that
. For a given value of
, if
, there exists an integer between
and
, and the cube root of this integer would be between
and
as desired. We seek the smallest
such that
.
Trying values of , we see that the smallest value of
that works is
.
Why is it and not greater than or equal to? - awesomediabrine
Because if its equal to, then there is no integer in between the two values. - resources
Solution 3 (Similar to Solution 2)
Since is less than
, we have
. Notice that since we want
minimized,
should also be minimized. Also,
should be as close as possible, but not exceeding
. This means
should be set to
. Substituting and simplifying, we get
The last two terms in the right side can be ignored in the calculation because they are too small. This results in
. The minimum positive integer
that satisfies this is
. ~ Hb10
Solution 4 (Calculus)
Note that the cube root is increasing for positive reals while its derivative is decreasing, so linear approximation gives
and
From this, it is clear that
is the smallest
for which LHS will be less than
. ~ Hyprox1413
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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