Difference between revisions of "2010 AIME I Problems/Problem 10"
(→Solution 3: Casework and Brute Force) |
|||
Line 18: | Line 18: | ||
We immediately see that <math>a_3</math> can only be <math>0</math>, <math>1</math> or <math>2</math>. We also note that the maximum possible value for <math>10a_1 + a_0</math> is <math>990 + 99 = 1089</math>. We then split into cases: | We immediately see that <math>a_3</math> can only be <math>0</math>, <math>1</math> or <math>2</math>. We also note that the maximum possible value for <math>10a_1 + a_0</math> is <math>990 + 99 = 1089</math>. We then split into cases: | ||
− | Case 1: <math>a_3 = 0</math> | + | Case 1: <math>a_3 = 0</math>. |
We try to find possible values of <math>a_2</math>. We plug in <math>a_3 = 0</math> and <math>10a_1 + a_0 = 1089</math> to our initial equation, which gives us <math>2010 = 0 + 100a_2^2 + 1089</math>. Thus <math>a_2 \geq 10</math>. We also see that <math>a_2 \leq 20</math>. We now take these values of <math>a_2</math> and find the number of pairs <math>(a_1, a_0)</math> that work. If <math>a_2 = 10</math>, <math>10a_1 + a_0 = 1010</math>. We see that there are <math>8</math> possible pairs in this case. Using the same logic, there are <math>10</math> ways for <math>a_2 = 11, 12 \ldots 19</math>. For <math>a_2 = 20</math>, we get the equation <math>10a_1 + a_0 = 10</math>, for 2 ways. Thus, for <math>a_3 = 0</math>, there are <math>8 + 10 \cdot 9 + 2 = 100</math> ways. | We try to find possible values of <math>a_2</math>. We plug in <math>a_3 = 0</math> and <math>10a_1 + a_0 = 1089</math> to our initial equation, which gives us <math>2010 = 0 + 100a_2^2 + 1089</math>. Thus <math>a_2 \geq 10</math>. We also see that <math>a_2 \leq 20</math>. We now take these values of <math>a_2</math> and find the number of pairs <math>(a_1, a_0)</math> that work. If <math>a_2 = 10</math>, <math>10a_1 + a_0 = 1010</math>. We see that there are <math>8</math> possible pairs in this case. Using the same logic, there are <math>10</math> ways for <math>a_2 = 11, 12 \ldots 19</math>. For <math>a_2 = 20</math>, we get the equation <math>10a_1 + a_0 = 10</math>, for 2 ways. Thus, for <math>a_3 = 0</math>, there are <math>8 + 10 \cdot 9 + 2 = 100</math> ways. | ||
− | Case 2: <math>a_3 = 1</math> | + | Case 2: <math>a_3 = 1</math>. |
This case is almost identical to the one above, except <math>0 \leq a_2 \leq 10</math>. We also get 100 ways. | This case is almost identical to the one above, except <math>0 \leq a_2 \leq 10</math>. We also get 100 ways. | ||
− | Case 3: <math>a_3 = 2</math> | + | Case 3: <math>a_3 = 2</math>. |
If <math>a_3 = 2</math>, our initial equation becomes <math>100a_2 + 10a_1 + a_0 = 10</math>. It is obvious that <math>a_2 = 0</math>, and we are left with <math>10a_1 + a_0 = 10</math>. We saw above that there are <math>2</math> ways. | If <math>a_3 = 2</math>, our initial equation becomes <math>100a_2 + 10a_1 + a_0 = 10</math>. It is obvious that <math>a_2 = 0</math>, and we are left with <math>10a_1 + a_0 = 10</math>. We saw above that there are <math>2</math> ways. | ||
Revision as of 17:15, 4 January 2016
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 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 3: 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.