The sum of the digits of a number N is 23. The remainder when N is divided by…

Solution:

• Sum of the digits is 23, so N ≡ 23 ≡ 5 (mod 9). Therefore N ≡ 5 (mod 9) ⇒ N ≡ 2 (mod 3).

• Given N ≡ 7 (mod 11).

Now solve the simultaneous congruences N ≡ 7 (mod 11) and N ≡ 2 (mod 3).

1. Write N = 11k + 7. Substitute into the mod 3 condition: 11k + 7 ≡ 2 (mod 3).

2. Reduce coefficients mod 3: 11 ≡ 2 (mod 3) and 7 ≡ 1 (mod 3), so 2k + 1 ≡ 2 (mod 3) ⇒ 2k ≡ 1 (mod 3).

3. Solve 2k ≡ 1 (mod 3). The inverse of 2 mod 3 is 2, so k ≡ 2 (mod 3). The smallest such k is 2, giving N = 11·2 + 7 = 29.

Thus N ≡ 29 (mod 33). Therefore the remainder when N is divided by 33 is 29.

Note: 29 is the least nonnegative representative of the congruence class; adding multiples of 33 gives all numbers satisfying both congruences.