Difference between revisions of "2010 AIME I Problems/Problem 10"
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~rzlng | ~rzlng | ||
+ | ==Solution 5== | ||
+ | First note that <math>a_3</math> has to be a single-digit number(<math>0</math>, <math>1</math>, or <math>2</math> to be exact), and that <math>a_1</math> has to be a two-digit multiple of ten. | ||
+ | Then, <math>a_3</math>, <math>a_2</math>, <math>a_1</math> and <math>a_0</math> can be represented as follows: | ||
+ | <cmath>\begin{align*} | ||
+ | a_3 = a | ||
+ | \\ a_2 = 10b+c | ||
+ | \\ a_1= 10d+e | ||
+ | \\ a_0 = 10f | ||
+ | \end{align*}</cmath> | ||
+ | , where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, <math>e</math>, and <math>f</math> are all(not necessarily nonzero) digits. | ||
+ | Now, we can write our given equation as follows: | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 2010 = 1000(a+b) + 100(c+d) + 10(e+f) | ||
+ | \\ 201 = 100(a+b) + 10(c+d) + (e+f) | ||
+ | \\ 201 = (100a + 10c + e) + (100b + d + f) | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Now, each integer between <math>0</math> and <math>201</math> inclusive can be represented in exactly one way as <math>100a + 10c + e</math>, and this corresponds with one unique <math>100b + d + f</math>, so it remains to count the number of integers between <math>0</math> and <math>201</math> inclusive. This is easily counted to be <math>\boxed{202}</math>. | ||
− | == Solution 6 | + | ==Solution 6 == |
− | + | Just note that this corresponds to <math>0 \leq 10^3 \cdot a_3 + 10^2 \cdot a_2 + 10^1 \cdot a_1 \leq 2010</math>, because we can use <math>a_0</math> to fill in the remaining gap. Then, dividing by <math>10</math>, we have <math>0 \leq \overline{a_3 a_2 a_1} \leq 201</math>, of which there are <math>\boxed{202}</math> solutions. | |
− | |||
− | |||
==Video Solution== | ==Video Solution== |
Latest revision as of 08:07, 15 August 2024
Contents
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 solutions is just the number of solutions to
where
, which is 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.
Solution 5: Generating Functions
We will represent the problem using generating functions. Consider the generating function
where the first factor represents
, the second factor
, and so forth. We want to find the coefficient of
in the expansion of
. Now rewriting each factor using the geometric series yields
The coefficient of
in this is simply
, as we can choose any of the first 202 terms from the second factor and pair it with exactly one term in the first factor.
~rzlng
Solution 5
First note that has to be a single-digit number(
,
, or
to be exact), and that
has to be a two-digit multiple of ten.
Then,
,
,
and
can be represented as follows:
, where
,
,
,
,
, and
are all(not necessarily nonzero) digits.
Now, we can write our given equation as follows:
Now, each integer between
and
inclusive can be represented in exactly one way as
, and this corresponds with one unique
, so it remains to count the number of integers between
and
inclusive. This is easily counted to be
.
Solution 6
Just note that this corresponds to , because we can use
to fill in the remaining gap. Then, dividing by
, we have
, of which there are
solutions.
Video Solution
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.