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Difference between revisions of "2012 AIME I Problems/Problem 1"

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== Solution ==
 
== Solution ==
  
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A positive integer is divisible by <math>4</math> if and only if its last two digits are divisible by <math>4.</math> For any value of <math>b</math>, there are two possible values for <math>a</math> and <math>c</math>, since we find that if <math>b</math> is even, <math>a</math> and <math>c</math> must be either <math>4</math> or <math>8</math>, and if <math>b</math> is odd, <math>a</math> and <math>c</math> must be either <math>2</math> or <math>6</math>. There are thus <math>2 * 2 = 4</math> ways to choose <math>a</math> and <math>c</math> for each <math>b,</math> and <math>10</math> ways to choose <math>b</math> since <math>b</math> can be any digit. The final answer is then <math>4 * 10 = \boxed{040}</math>.
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A positive integer is divisible by <math>4</math> if and only if its last two digits are divisible by <math>4.</math> For any value of <math>b</math>, there are two possible values for <math>a</math> and <math>c</math>, since we find that if <math>b</math> is even, <math>a</math> and <math>c</math> must be either <math>4</math> or <math>8</math>, and if <math>b</math> is odd, <math>a</math> and <math>c</math> must be either <math>2</math> or <math>6</math>. There are thus <math>2 \cdot 2 = 4</math> ways to choose <math>a</math> and <math>c</math> for each <math>b,</math> and <math>10</math> ways to choose <math>b</math> since <math>b</math> can be any digit. The final answer is then <math>4 \cdot 10 = \boxed{040}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|before=First Problem|num-a=2}}
 
{{AIME box|year=2012|n=I|before=First Problem|num-a=2}}

Revision as of 00:48, 17 March 2012

Problem 1

Find the number of positive integers with three not necessarily distinct digits, $abc$, with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$.

Solution

A positive integer is divisible by $4$ if and only if its last two digits are divisible by $4.$ For any value of $b$, there are two possible values for $a$ and $c$, since we find that if $b$ is even, $a$ and $c$ must be either $4$ or $8$, and if $b$ is odd, $a$ and $c$ must be either $2$ or $6$. There are thus $2 \cdot 2 = 4$ ways to choose $a$ and $c$ for each $b,$ and $10$ ways to choose $b$ since $b$ can be any digit. The final answer is then $4 \cdot 10 = \boxed{040}$.

See also

2012 AIME I (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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