Find the divisor from two remainder statements A number leaves remainder 24 when divided by a certain divisor. Twice the number leaves remainder 11 with the same divisor. Determine the divisor.

Introduction / Context: Problems that provide remainders for N and for 2N invite the use of modular equivalences. By translating each sentence into a congruence and comparing them, we can deduce the divisor without ever computing N itself.

Given Data / Assumptions:

• N ≡ 24 (mod d).
• 2N ≡ 11 (mod d).
• d is a positive integer greater than the remainders where applicable.

Concept / Approach: From N ≡ 24 (mod d), doubling gives 2N ≡ 48 (mod d). But we also have 2N ≡ 11 (mod d). Therefore, 48 ≡ 11 (mod d) which implies d divides (48 − 11) = 37. Since a positive divisor greater than the remainder 24 is required, the only valid d is 37.

Step-by-Step Solution:Start: N ≡ 24 (mod d).Double: 2N ≡ 48 (mod d).Compare with given: 2N ≡ 11 (mod d) ⇒ 48 ≡ 11 (mod d).Hence d | (48 − 11) = 37 ⇒ d = 37 (since d > 24).

Verification / Alternative check: Construct N = 37k + 24. Then 2N = 74k + 48. Mod 37, 2N ≡ 11 because 48 mod 37 = 11, consistent with the condition for any integer k.

Why Other Options Are Wrong:13, 59, 35 do not divide 37. Only 37 satisfies 48 ≡ 11 (mod d) and is larger than 24.

Common Pitfalls: Forgetting to double the first congruence; assuming d equals the difference between remainders without checking divisibility; overlooking the constraint that remainders must be less than the divisor.

Final Answer: 37