2012 AIME II Problems/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.
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.
Solution 2 (Probability)
Note that and . Now, consider any positive integer such that . Since all -safe numbers are of the form or , the probability that is -safe is . Similarly, the probability that is -safe is and the probability that is -safe is . Furthermore, since are all relatively prime, being -safe, -safe, and -safe are all independent events. Thus, the number of positive integers less then or equal to which are simultaneously -safe, -safe, and -safe is . However, the problem asks for numbers less than or equal to , so we must subtract any numbers we counted which are greater than . This is easy; we can see that and were counted, so the answer is . -brainiacmaniac31
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