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
Solution 1
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 less than or equal to
.
In summary, substituting and
we finally find that
.
Solution 2
We know that once we have found the largest value of , all values after
are less than
. Therefore, we are looking for the largest possible value such that:
Dividing by gives:
.
We can express these binomial coefficients as factorials.
We note that the can cancel. Also,
. Similarly,
.
Canceling these terms yields,
Cross multiplying gives:
Therefore, since this identity holds for all values of , the largest possible value of
is
.
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 |
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