2010 AIME I Problems/Problem 10
Contents
Problem
Problem
Let be the number of ways to write
in the form
, where the
's are integers, and
. An example of such a representation is
. Find
.
Solution 1
If we choose and
such that
there is a unique choice of
and
that makes the equality hold. So
is just the number of combinations of
and
we can pick. If
or
we can let
be anything from
to
. If
then
or
. Thus
.
Solution 2
Note that is the base
representation of any number from
to
, and similarly
is ten times the base
representation of any number from
to
. Thus, the number of solution is just the number of solutions to
where
, which is clearly equal to
as
can range from
to
.
Solution 3
Note that and
. It's easy to see that exactly 10 values in
that satisfy our first congruence. Similarly, there are 10 possible values of
for each choice of
. Thus, there are
possible choices for
and
. We next note that if
and
are chosen, then a valid value of
determines
, so we dive into some simple casework:
- If
, there are 3 valid choices for
. There are only 2 possible cases where
, namely
. Thus, there are
possible representations in this case.
- If
,
can only equal 0. However, this case cannot occur, as
. Thus,
. However,
. Thus, we have
always.
- If
, then there are 2 valid choices for
. Since there are 100 possible choices for
and
, and we have already checked the other cases, it follows that
choices of
and
fall under this case. Thus, there are
possible representations in this case.
Our answer is thus .
Solution 4: Casework and Brute Force
We immediately see that can only be
,
or
. We also note that the maximum possible value for
is
. We then split into cases:
Case 1: .
We try to find possible values of
. We plug in
and
to our initial equation, which gives us
. Thus
. We also see that
. We now take these values of
and find the number of pairs
that work. If
,
. We see that there are
possible pairs in this case. Using the same logic, there are
ways for
. For
, we get the equation
, for 2 ways. Thus, for
, there are
ways.
Case 2: .
This case is almost identical to the one above, except
. We also get 100 ways.
Case 3: .
If
, our initial equation becomes
. It is obvious that
, and we are left with
. We saw above that there are
ways.
Totaling everything, we get that there are ways.
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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