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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===
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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.
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*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.
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*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.
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Our answer is thus <math>6 + 0 + 196 = \boxed{202}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 13:00, 18 March 2015

Problem

Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$, where the $a_i$'s are integers, and $0 \le a_i \le 99$. An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$. Find $N$.

Solution

Solution 1

If we choose $a_3$ and $a_1$ such that $(10^3)(a_3) + (10)(a_1) \leq 2010$ there is a unique choice of $a_2$ and $a_0$ that makes the equality hold. So $N$ is just the number of combinations of $a_3$ and $a_1$ we can pick. If $a_3 = 0$ or $a_3 = 1$ we can let $a_1$ be anything from $0$ to $99$. If $a_3 = 2$ then $a_1 = 0$ or $a_1 = 1$. Thus $N = 100 + 100 + 2 = \fbox{202}$.

Solution 2

Note that $a_0 \equiv 2010\ (\textrm{mod}\ 10)$ and $a_1 \equiv 2010 - a_0\ (\textrm{mod}\ 100)$. It's easy to see that exactly 10 values in $0 \leq a_0 \leq 99$ that satisfy our first congruence. Similarly, there are 10 possible values of $a_1$ for each choice of $a_0$. Thus, there are $10 \times 10 = 100$ possible choices for $a_0$ and $a_1$. We next note that if $a_0$ and $a_1$ are chosen, then a valid value of $a_3$ determines $a_2$, so we dive into some simple casework:

  • If $2010 - 10a_1 - a_0 \geq 2000$, there are 3 valid choices for $a_3$. There are only 2 possible cases where $2010 - 10a_1 - a_0 \geq 2000$, namely $(a_1, a_0) = (1,0), (10,0)$. Thus, there are $3 \times 2 = 6$ possible representations in this case.
  • If $2010 - 10a_1 - a_0 < 1000$, $a_3$ can only equal 0. However, this case cannot occur, as $10a_1+a_0\leq 990+99 = 1089$. Thus, $2010-10a_1-a_0 \geq 921$. However, $2010-10a_1-a_0 = 1000a_3 + 100a_2 \equiv 0\ (\textrm{mod}\ 100)$. Thus, we have $2010-10a_1-a_0 \geq 1000$ always.
  • If $1000 \leq 2010 - 10a_1 - a_0 < 2000$, then there are 2 valid choices for $a_3$. Since there are 100 possible choices for $a_0$ and $a_1$, and we have already checked the other cases, it follows that $100 - 2 - 0 = 98$ choices of $a_0$ and $a_1$ fall under this case. Thus, there are $2 \times 98 = 196$ possible representations in this case.

Our answer is thus $6 + 0 + 196 = \boxed{202}$.

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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