1991 AIME Problems/Problem 3
Problem
Expanding by the binomial theorem and doing no further manipulation gives
where
for
. For which
is
the largest?
Solution
Let . Then we may write
. Taking logarithms in both sides of this last equation and using the well-known fact
(valid if
), we have
Now, keeps increasing with
as long as the arguments
in each of the terms (recall that
if
). Therefore, the integer
that we are looking for must satisfy
, where
denotes the greatest integer smaller than or equal to
.
In summary, substituting and
we finally find that
.
See also
1991 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |