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

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== Problem 1 ==
 
== Problem 1 ==
 
Find the number of positive integers with three not necessarily distinct digits, <math>abc</math>, with <math>a \neq 0</math> and <math>c \neq 0</math> such that both <math>abc</math> and <math>cba</math> are multiples of <math>4</math>.
 
Find the number of positive integers with three not necessarily distinct digits, <math>abc</math>, with <math>a \neq 0</math> and <math>c \neq 0</math> such that both <math>abc</math> and <math>cba</math> are multiples of <math>4</math>.
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== 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 4. 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>. Thus, there are <math>2 * 2 = 4</math> ways to choose <math>a</math> and <math>c</math>, and <math>10</math> ways to choose <math>b</math> (since <math>b</math> can be any digit). Therefore, the final answer is <math>4 * 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 23:19, 16 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 $6$. Thus, there are $2 * 2 = 4$ ways to choose $a$ and $c$, and $10$ ways to choose $b$ (since $b$ can be any digit). Therefore, the final answer is $4 * 10 = \boxed{040}$.

See also

2012 AIME I (ProblemsAnswer KeyResources)
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