2013 Mock AIME I Problems/Problem 6
Contents
[hide]Problem 6
Find the number of integer values can have such that the equation
has a solution.
Solution 1
is a continuous function, so every value between its minimum and maximum is attainable by the Intermediate Value Theorem. By Cauchy-Schwarz,
Giving a maximum of
, which is achievable when
. Note that a minimum of
can be attained at
. Thus the values of
that work are the integers from
to
, inclusive, giving a total of
.
Solution 2 (calculus)
As in the first solution, let . Then,
. Thus,
has maxima and minima when
. After squaring both sides and applying the Pythagorean Identity, we solve for
:
, so
. The maximum of the function occurs when
, so the maximum is
. Likewise, when both
and
are negative, the minimum
occurs. We can now proceed as in the first solution to solve the problem to get our answer
.