1985 AHSME Problems/Problem 30
Problem
Let be the greatest integer less than or equal to
. Then the number of real solutions to
is
Solution
We can rearrange the equation into . Obviously, the RHS is an integer, so
for some integer
. We can therefore make the substitution
to get
(We'll try the case where later.) Now let
for a nonnegative integer
, so that
.
Going back to , this implies
. Making the substitution
gives the system of inequalities
The first inequality simplifies to , or
. Since
is even, we must have
, so
. Therefore,
. The second inequality simplifies to
, or
. Therefore, as
is even, we have
, or
. Hence
or
Since both inequalities must be satisfied, we see that only
,
,
, and
satisfy both inequalities. Each will lead to a distinct solution for
, and thus for
, for a total of
positive solutions.
Now let . We have
Since , this can be rewritten as
Since is positive, the least possible value of
is
, hence
, i.e., it must be negative. But it is also equal to
, which is positive, and this is a contradiction. Therefore, there are no negative roots.
The total number of roots to this equation is thus , or
.
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
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