2012 AIME I Problems/Problem 1

Problem

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 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 Solutions

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

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

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

Video Solution

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

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

Page

Toolbox

Search