Difference between revisions of "2010 AIME I Problems/Problem 10"
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== Solution == | == Solution == | ||
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+ | ===Solution 1=== | ||
If we choose <math>a_3</math> and <math>a_1</math> such that <math>(10^3)(a_3) + (10)(a_1) \leq 2010</math> there is a [[bijection|unique]] choice of <math>a_2</math> and <math>a_0</math> that makes the equality hold. So <math>N</math> is just the number of combinations of <math>a_3</math> and <math>a_1</math> we can pick. If <math>a_3 = 0</math> or <math>a_3 = 1</math> we can let <math>a_1</math> be anything from <math>0</math> to <math>99</math>. If <math>a_3 = 2</math> then <math>a_1 = 0</math> or <math>a_1 = 1</math>. Thus <math>N = 100 + 100 + 2 = \fbox{202}</math>. | If we choose <math>a_3</math> and <math>a_1</math> such that <math>(10^3)(a_3) + (10)(a_1) \leq 2010</math> there is a [[bijection|unique]] choice of <math>a_2</math> and <math>a_0</math> that makes the equality hold. So <math>N</math> is just the number of combinations of <math>a_3</math> and <math>a_1</math> we can pick. If <math>a_3 = 0</math> or <math>a_3 = 1</math> we can let <math>a_1</math> be anything from <math>0</math> to <math>99</math>. If <math>a_3 = 2</math> then <math>a_1 = 0</math> or <math>a_1 = 1</math>. Thus <math>N = 100 + 100 + 2 = \fbox{202}</math>. | ||
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+ | ===Solution 2=== | ||
+ | |||
+ | Note that <math>a_0 \equiv 2010\ (\textrm{mod}\ 10)</math> and <math>a_1 \equiv 2010 - a_0\ (\textrm{mod}\ 100)</math>. It's easy to see that exactly 10 values in <math>0 \leq a_0 \leq 99</math> that satisfy our first congruence. Similarly, there are 10 possible values of <math>a_1</math> for each choice of <math>a_0</math>. Thus, there are <math>10 \times 10 = 100</math> possible choices for <math>a_0</math> and <math>a_1</math>. We next note that if <math>a_0</math> and <math>a_1</math> are chosen, then a valid value of <math>a_3</math> determines <math>a_2</math>, so we dive into some simple casework: | ||
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+ | *If <math>2010 - 10a_1 - a_0 \geq 2000</math>, there are 3 valid choices for <math>a_3</math>. There are only 2 possible cases where <math>2010 - 10a_1 - a_0 \geq 2000</math>, namely <math>(a_1, a_0) = (1,0), (10,0)</math>. Thus, there are <math>3 \times 2 = 6</math> possible representations in this case. | ||
+ | *If <math>2010 - 10a_1 - a_0 < 1000</math>, <math>a_3</math> can only equal 0. However, this case cannot occur, as <math>10a_1+a_0\leq 990+99 = 1089</math>. Thus, <math>2010-10a_1-a_0 \geq 921</math>. However, <math>2010-10a_1-a_0 = 1000a_3 + 100a_2 \equiv 0\ (\textrm{mod}\ 100)</math>. Thus, we have <math>2010-10a_1-a_0 \geq 1000</math> always. | ||
+ | *If <math>1000 \leq 2010 - 10a_1 - a_0 < 2000</math>, then there are 2 valid choices for <math>a_3</math>. Since there are 100 possible choices for <math>a_0</math> and <math>a_1</math>, and we have already checked the other cases, it follows that <math>100 - 2 - 0 = 98</math> choices of <math>a_0</math> and <math>a_1</math> fall under this case. Thus, there are <math>2 \times 98 = 196</math> possible representations in this case. | ||
+ | |||
+ | Our answer is thus <math>6 + 0 + 196 = \boxed{202}</math>. | ||
== See Also == | == See Also == |
Revision as of 14:00, 18 March 2015
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
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 .
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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