2012 AIME I Problems/Problem 1
Contents
Problem 1
Find the number of positive integers with three not necessarily distinct digits, , with and such that both and are multiples of .
Solutions
Solution 1
A positive integer is divisible by if and only if its last two digits are divisible by For any value of , there are two possible values for and , since we find that if is even, and must be either or , and if is odd, and must be either or . There are thus ways to choose and for each and ways to choose since can be any digit. The final answer is then .
Solution 2
A number is divisible by four if its last two digits are divisible by 4. Thus, we require that and are both divisible by . If is odd, then and must both be meaning that and are or . If is even, then and must be meaning that and are or . For each choice of there are choices for and for for a total of numbers.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/289
~ dolphin7
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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