Smallest number with specified remainders: Find the least positive number that leaves remainders 3, 5, and 8 when divided by 16, 18, and 21 respectively.

Difficulty: Medium

Correct Answer: 995

Explanation:


Introduction / Context:
This is a Chinese Remainder style problem with a useful observation: the three remainders (3, 5, 8) are all congruent to −13 with respect to their moduli (16, 18, 21). That means the sought number differs by the same constant from a common multiple of the moduli.



Given Data / Assumptions:

  • N ≡ 3 (mod 16)
  • N ≡ 5 (mod 18)
  • N ≡ 8 (mod 21)


Concept / Approach:
Note that 3 = −13 mod 16, 5 = −13 mod 18, 8 = −13 mod 21. Therefore, N ≡ −13 modulo LCM(16, 18, 21). Compute the LCM and then find the least positive N of the form LCM − 13.



Step-by-Step Solution:

LCM(16, 18, 21): 16 = 2^4, 18 = 2 * 3^2, 21 = 3 * 7 ⇒ LCM = 2^4 * 3^2 * 7 = 1008Thus N ≡ −13 (mod 1008) ⇒ least positive N = 1008 − 13 = 995Hence, the smallest such number is 995.


Verification / Alternative check:
Check: 995 mod 16 = 3, 995 mod 18 = 5, 995 mod 21 = 8. All conditions satisfied.



Why Other Options Are Wrong:
893, 992, 1024, and 1005 do not satisfy all three remainder conditions simultaneously.



Common Pitfalls:
Attempting to solve each congruence independently without noticing the convenient common shift of −13, or computing LCM incorrectly.



Final Answer:
995

More Questions from Problems on H.C.F and L.C.M

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