Greatest four-digit multiple: Find the greatest four-digit number that is divisible by each of 12, 18, 21, and 28.

Difficulty: Easy

Correct Answer: 9828

Explanation:


Introduction / Context:
The largest four-digit number divisible by several integers must be a multiple of their least common multiple (LCM). Compute the LCM, then take the largest multiple not exceeding 9999.



Given Data / Assumptions:

  • Divisors: 12, 18, 21, 28
  • Upper limit: 9999 (largest four-digit number)


Concept / Approach:
LCM via prime powers: choose the highest power of each prime appearing in the prime factorizations. Then compute floor(9999 / LCM) * LCM.



Step-by-Step Solution:

Prime factors: 12 = 2^2 * 3; 18 = 2 * 3^2; 21 = 3 * 7; 28 = 2^2 * 7LCM = 2^2 * 3^2 * 7 = 4 * 9 * 7 = 252Largest multiple ≤ 9999: floor(9999 / 252) = 39 ⇒ 39 * 252 = 9828Therefore, the greatest four-digit number divisible by all four is 9828.


Verification / Alternative check:
Check divisibility: 9828 / 12 = 819, /18 = 546, /21 = 468, /28 = 351 — all integers.



Why Other Options Are Wrong:
9864 and 9848 are not multiples of 252; 9636 is a multiple of 252 but smaller than 9828; 9996 is not divisible by all listed numbers.



Common Pitfalls:
Computing LCM incorrectly (e.g., missing 3^2) or multiplying by 40 (which gives 10080, exceeding four digits).



Final Answer:
9828

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

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