A number theory problem by Akshat Sharda

We are looking for numbers that are NOT equal to $0, \pm 1, \pm 2$ (mod 7, 11, or 13).

Put another way, we want numbers that are simultaneously:

3 or 4 (mod 7);

3, 4, 5, 6, 7, or 8 (mod 11); and

3, 4, 5, 6, 7, 8, 9, or 10 (mod 13).

There are 2 * 6 * 8 = 96 possible combinations of the restrictions above.

By the Chinese Remainder Theorem, since 7, 11, and 13 are relatively prime, it follows that each of these combinations yields a unique solution (mod 7 * 11 * 13 = 1001). That is, there are 96 unique numbers between 1 and 1001 (inclusive) that are 7-safe, 11-safe, and 13-safe.

Since any set of numbers $\{i \in \mathbb{Z} \mid n \leq i \leq n + 1000, n \in \mathbb{Z} \}$ is a complete residue system (mod 1001), it follows that there are 96 more unique solutions for every adjacent block of 1001 numbers. Specifically, there are 960 unique solutions less than or equal to 10010.

It's easy to manually check that the only simultaneously 7-safe, 11-safe and 13-safe numbers in the range $\{i \mid 10001 \leq i \leq 10010\}$ are 10006 and 10007, so we've over-counted by two, giving a final answer of 958 solutions.

Great question! $$