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

Revision as of 22:26, 17 January 2021

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 $4$.

Solutions

Solution 1

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 $4 \cdot 10 = \boxed{040}$.

Solution 2

A number is divisible by four if its last two digits are divisible by 4. Thus, we require that $10b + a$ and $10b + c$ are both divisible by $4$. If $b$ is odd, then $a$ and $c$ must both be $2 \pmod 4$ meaning that $a$ and $c$ are $2$ or $6$. If $b$ is even, then $a$ and $c$ must be $0 \pmod 4$ meaning that $a$ and $c$ are $4$ or $8$. For each choice of $b$ there are $2$ choices for $a$ and $2$ for $c$ for a total of $10 \cdot 2 \cdot 2 = \boxed{040}$ numbers.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012aimei/289

~ dolphin7

Video Solution

https://youtu.be/ZhAZ1oPe5Ds?t=3235

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=T8Ox412AkZc ~Shreyas S

See also

2012 AIME I (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
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All AIME Problems and Solutions

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