2012 AIME II Problems/Problem 12

Problem 12

For a positive integer $p$ , define the positive integer $n$ to be $p$ -safe if $n$ differs in absolute value by more than $2$ from all multiples of $p$ . For example, the set of $10$ -safe numbers is $\{ 3, 4, 5, 6, 7, 13, 14, 15, 16, 17, 23, \ldots\}$ . Find the number of positive integers less than or equal to $10,000$ which are simultaneously $7$ -safe, $11$ -safe, and $13$ -safe.

Solution

We see that a number $n$ is $p$ -safe if and only if the residue of $n \mod p$ is greater than $2$ and less than $p-2$ ; thus, there are $p-5$ residues $\mod p$ that a $p$ -safe number can have. Therefore, a number $n$ satisfying the conditions of the problem can have $2$ different residues $\mod 7$ , $6$ different residues $\mod 11$ , and $8$ different residues $\mod 13$ . The Chinese Remainder Theorem states that for a number $x$ that is $a (mod b)$ $c (mod d)$ $e (mod f)$ has one solution if $gcd(b,d,f)=1$ . For example, in our case, the number $n$ can be: $3 (mod 7)$ $3 (mod 11)$ $7 (mod 13)$ so since gcd $(7,11,13)$ =1, there is 1 solution for n for this case of residues of $n$ .

This means that by the Chinese Remainder Theorem, $n$ can have $2\cdot 6 \cdot 8 = 96$ different residues mod $7 \cdot 11 \cdot 13 = 1001$ . Thus, there are $960$ values of $n$ satisfying the conditions in the range $0 \le n < 10010$ . However, we must now remove any values greater than $10000$ that satisfy the conditions. By checking residues, we easily see that the only such values are $10007$ and $10006$ , so there remain $\fbox{958}$ values satisfying the conditions of the problem.

2012 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2012 AIME II Problems/Problem 12

Problem 12

Solution

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