Find the remainder when 83 × 85 × 87 × 88 × 91 is divided by 41.

When a product is divided by a number n, each factor can first be reduced modulo n; the remainder of the original product by n equals the remainder of the product of these reduced values by n. This holds because congruence is preserved under multiplication — if a ≡ a′ (mod n) and b ≡ b′ (mod n), then a × b ≡ a′ × b′ (mod n).

Applying this to 83 × 85 × 87 × 88 × 91 divided by 41:

1. 83 = 2 × 41 + 1, so 83 ≡ 1 (mod 41).

2. 85 = 2 × 41 + 3, so 85 ≡ 3 (mod 41).

3. 87 = 2 × 41 + 5, so 87 ≡ 5 (mod 41).

4. 88 = 2 × 41 + 6, so 88 ≡ 6 (mod 41).

5. 91 = 2 × 41 + 9, so 91 ≡ 9 (mod 41).

6. Multiplying the reduced values: 1 × 3 × 5 × 6 × 9 = 810.

7. Dividing 810 by 41: 41 × 19 = 779, and 810 − 779 = 31, so 810 ≡ 31 (mod 41).

Cross-check with a different grouping order: 6 × 9 = 54 ≡ 13 (mod 41); 13 × 5 = 65 ≡ 24 (mod 41); 24 × 3 = 72 ≡ 31 (mod 41); 31 × 1 = 31. The same remainder confirms the result.

Therefore, the remainder when 83 × 85 × 87 × 88 × 91 is divided by 41 is 31.