Find the greatest integer that divides each of the numbers 263, 935, and 1383 and leaves a remainder of 7 in every case.
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A30
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B31
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C32
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D35
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E28
Answer
Correct Answer: 32
Explanation
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:
Transform numbers: 263 − 7 = 256; 935 − 7 = 928; 1383 − 7 = 1376.Compute gcd(256, 928): 928 mod 256 = 160; 256 mod 160 = 96; 160 mod 96 = 64; 96 mod 64 = 32; 64 mod 32 = 0 ⇒ gcd = 32.Now gcd(32, 1376): 1376 mod 32 = 0 ⇒ gcd = 32.Therefore, the greatest required divisor is 32.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