Find the greatest integer that divides each of the numbers 263, 935, and 1383 and leaves a remainder of 7 in every case.

Introduction / Context:
This divisibility problem asks for the largest integer d that, when dividing 263, 935, and 1383, leaves the same remainder 7 in each case. Such questions reduce to computing a greatest common divisor (GCD) after adjusting for the common remainder.

Given Data / Assumptions:

• The three integers are 263, 935, and 1383.
• The common remainder on division by the required number is 7.
• d is a positive integer and must be as large as possible.

Concept / Approach:
If each number leaves remainder 7 upon division by d, then subtracting 7 from each number yields exact multiples of d. Therefore, d must divide 263 − 7, 935 − 7, and 1383 − 7. Hence, d = gcd(256, 928, 1376). Compute the GCD stepwise using the Euclidean algorithm.

Step-by-Step Solution:

Verification / Alternative check:
Direct check: 256/32, 928/32, and 1376/32 are integers (8, 29, and 43). This confirms 32 divides all adjusted values, and no larger divisor can exceed the GCD.

Why Other Options Are Wrong:

• 31, 35, 30, 28: None divides all of 256, 928, and 1376 exactly, so they cannot be the common divisor after adjusting the remainder.

Common Pitfalls:

• Forgetting to subtract the common remainder before computing the GCD.
• Stopping the Euclidean algorithm too early and reporting a non-maximum common divisor.

Final Answer:
32