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Difference between revisions of "2010 AIME I Problems/Problem 10"

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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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== See also ==
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== See Also ==
 
{{AIME box|year=2010|num-b=9|num-a=11|n=I}}
 
{{AIME box|year=2010|num-b=9|num-a=11|n=I}}
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 15:55, 12 April 2012

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

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 $N = 100 + 100 + 2 = \fbox{202}$.

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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