2012 AIME I Problems/Problem 1
Contents
Problem
Find the number of positive integers with three not necessarily distinct digits, , with and such that both and are multiples of .
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.
Solution 3
For this number to fit the requirements and must be divisible by 4. So and so must for each two digits of . There are two possibilities for if is odd and three possibilities if is even. So there are possibilities but this overcounts when or . So when and the corresponding should be removed, so . But we are still overcounting when is even because then can be 0. So the answer is
~LuisFonseca123
Video Solution by OmegaLearn
https://youtu.be/ZhAZ1oPe5Ds?t=3235
~ pi_is_3.14
Video Solutions
https://artofproblemsolving.com/videos/amc/2012aimei/289
https://www.youtube.com/watch?v=T8Ox412AkZc
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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