Problem:
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:
Because , the Chinese Remainder Theorem implies that there is a one-to-one correspondence between the integers from to and the ordered triples , where the number corresponds to the triple if the remainders when is divided by and are , and , respectively. The number is -safe if , -safe if , and -safe if . Thus there are positive integers less than that are -safe, -safe, and -safe. The same can be said for the integers in the range from to , those from to , those from to , and so forth. It follows that there are numbers between and that are simultaneously -safe, -safe, and -safe. The numbers and are the only integers between and that are -safe, -safe, and -safe, so the required number is .
The problems on this page are the property of the MAA's American Mathematics Competitions