How can you test 7, 11, and 13 at once using 1001?

Group the digits into 3-digit blocks from the right and take the alternating sum of the blocks — because 1001 = 7 × 11 × 13.

Since 1000 ≡ −1 (mod 7, 11, and 13), blocks alternate sign just like single digits do for 11:

• 4,004 → blocks 004 and 4 → 4 − 4 = 0 → divisible by 7, 11, and 13. ✓ (4004 = 4 × 7 × 11 × 13)
• A number like 1,234,567 → blocks 567, 234, 1 → 567 − 234 + 1 = 334 → check 334 against 7/11/13.

One alternating block sum simultaneously gives the remainder for all three primes.