2013 AIME I Problems/Problem 8
Contents
Problem 8
The domain of the function is a closed interval of length
, where
and
are positive integers and
. Find the remainder when the smallest possible sum
is divided by 1000.
Solution
We know that the domain of is
, so
. Now we can apply the definition of logarithms:
Since the domain of
has length
, we have that
A larger value of will also result in a larger value of
since the numerator increases faster than the denominator, so we want to find the smallest value of
that also results in an integer value of
. The problem states that
. Thus, first we try
:
Now, we try
:
Since
is the smallest value of
that results in an integral
value, we have minimized
, which is
.
Solution 2
We start with the same method as above. The domain of the arcsin function is , so
.
For to be an integer,
must divide
, and
. To minimize
,
should be as small as possible because increasing
will decrease
, the amount you are subtracting, and increase
, the amount you are adding; this also leads to a small
which clearly minimizes
.
We let equal
, the smallest factor of
that isn't
. Then we have
, so the answer is
.
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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