Number theory and least common multiple (LCM) remainders puzzle I have packs of sweets. If I put 2, 3, or 4 sweets per pack, I have 1 sweet left over each time. If I pack 5 per pack, none are left. What is the minimum number of sweets?

Introduction / Context:
This is a classic modular arithmetic and least common multiple question. You must find the smallest positive integer that fits several remainder conditions simultaneously, which is common in aptitude tests and coding interviews.

Given Data / Assumptions:

• N ≡ 1 (mod 2).
• N ≡ 1 (mod 3).
• N ≡ 1 (mod 4).
• N ≡ 0 (mod 5).

Concept / Approach:
Combine the first three congruences using the least common multiple. If N leaves remainder 1 when divided by 2, 3, and 4, then N = 12k + 1 because lcm(2, 3, 4) = 12. Then enforce the multiple-of-5 condition to find the smallest k that satisfies all requirements.

Step-by-Step Solution:

Verification / Alternative check:
25 divided by 2, 3, or 4 leaves remainder 1; 25 divided by 5 leaves remainder 0. All constraints check out.

Why Other Options Are Wrong:

• 37, 54, 65: At least one of the given modular conditions fails for each of these numbers.

Common Pitfalls:
Forgetting to use lcm for the combined remainder-1 constraints or misapplying the modular equivalence when reducing 12 modulo 5.

Final Answer:
25