Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12

General
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determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

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Determine the smallest 3-digit number which is exactly divisible by 6, 8, and 12.

Answer: To find the smallest 3-digit number which is exactly divisible by 6, 8, and 12, follow these steps:

  1. Find the Least Common Multiple (LCM) of 6, 8, and 12:

    • First, find the prime factorizations of 6, 8, and 12.

      6 = 2^1 \times 3^1
      8 = 2^3
      12 = 2^2 \times 3^1

    • Then, determine the LCM by taking the highest power of each prime that appears in the factorizations.

      \text{LCM} = 2^3 \times 3^1

    • Calculate the LCM:

      2^3 = 8
      3^1 = 3
      \text{LCM} = 8 \times 3 = 24

  2. Identify the smallest 3-digit number:

    • The smallest 3-digit number is 100.
    • We need to find the smallest 3-digit number that is a multiple of 24.

  3. Divide 100 by 24 and round up to the next integer:

    • Calculate \frac{100}{24}:

      \frac{100}{24} \approx 4.1667

    • Round up to the next integer to get the smallest integer k:

      k = \lceil 4.1667 \rceil = 5

  4. Multiply the LCM by this smallest integer k:

    • Calculate 24 \times 5:

      24 \times 5 = 120

Therefore, the smallest 3-digit number which is exactly divisible by 6, 8, and 12 is \boxed{120}.