Difficulty: Medium
Correct Answer: 358
Explanation:
Introduction / Context:
This is a classic modular arithmetic problem involving remainders. We are given two different group sizes (25 and 28) and the remainders of sweets left over in each case. The task is to determine the total number of sweets consistent with both conditions.
Given Data / Assumptions:
Concept / Approach:
Use modular arithmetic. The conditions translate to two congruences:
S mod 25 = 8S mod 28 = 22We can solve by inspection, systematic search, or using the Chinese Remainder approach because 25 and 28 are coprime (gcd = 1).
Step-by-Step Solution:
1) From S mod 25 = 8 ⇒ S = 25k + 8 for some integer k.2) Substitute into the second condition: 25k + 8 ≡ 22 (mod 28) ⇒ 25k ≡ 14 (mod 28).3) Reduce 25k mod 28: since 25 ≡ -3 (mod 28), we have -3k ≡ 14 (mod 28) ⇒ 3k ≡ -14 ≡ 14 (mod 28).4) Solve 3k ≡ 14 (mod 28). The inverse of 3 mod 28 is 19 (since 319 = 57 ≡ 1 mod 28). Thus k ≡ 1419 = 266 ≡ 266 - 928 = 266 - 252 = 14 (mod 28).5) Take k = 14 (smallest positive). Then S = 2514 + 8 = 350 + 8 = 358.
Verification / Alternative check:
Check S mod 25: 358 = 2514 + 8 ⇒ remainder 8 ✔Check S mod 28: 358 = 2812 + 22 ⇒ remainder 22 ✔Thus S = 358 fits both conditions.
Why Other Options Are Wrong:
Common Pitfalls:
Subtracting the remainders directly or guessing. Always translate to precise congruences and solve systematically to avoid errors.
Final Answer:
358
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