2012 AIME II Problems/Problem 12
Problem 12
For a positive integer , define the positive integer to be -safe if differs in absolute value by more than from all multiples of . For example, the set of -safe numbers is . Find the number of positive integers less than or equal to which are simultaneously -safe, -safe, and -safe.
Solution
We see that a number is -safe if and only if the residue of is greater than and less than ; thus, there are residues that a -safe number can have. Therefore, a number satisfying the conditions of the problem can have different residues , different residues , and different residues . The Chinese Remainder Theorem states that for a number that is (mod b) (mod d) (mod f) has one solution if . For example, in our case, the number can be: 3 (mod 7) 3 (mod 11) 7 (mod 13) so since =1, there is 1 solution for n for this case of residues of .
This means that by the Chinese Remainder Theorem, can have different residues mod . Thus, there are values of satisfying the conditions in the range . However, we must now remove any values greater than that satisfy the conditions. By checking residues, we easily see that the only such values are and , so there remain values satisfying the conditions of the problem.
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 13 | |
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