Question

The number of factors of 1800 that are multiple of 6 is …………. .

Hint:

For counting the number of factors of a number, first find its prime factorization. Then, for each prime factor, the exponent can range from 0 to the exponent in the factorization. Multiply the number of choices for each prime factor.

Answer 1

1. Prime Factorization of 1800:

   1800 = 18 × 100

   1800 = 2 × 9 × 10 × 10

   1800 = 2 × 3 × 3 × 2 × 5 × 2 × 5

   1800 = 2³ × 3² × 5²

2. Factors of 1800:

   Any factor of 1800 will have the form 2^a × 3^b × 5^c, where:

   • 0 ≤ a ≤ 3
   • 0 ≤ b ≤ 2
   • 0 ≤ c ≤ 2
3. Multiples of 6:

   For a factor to be a multiple of 6, it must be divisible by 6. Since 6 = 2 × 3, the factor must have at least one 2 and one 3 in its prime factorization. This means:

   • 1 ≤ a ≤ 3 (a can be 1, 2, or 3)
   • 1 ≤ b ≤ 2 (b can be 1 or 2)
   • 0 ≤ c ≤ 2 (c can be 0, 1, or 2)
4. Counting the Possibilities:
   • Choices for a: 3 (1, 2, or 3)
   • Choices for b: 2 (1 or 2)
   • Choices for c: 3 (0, 1, or 2)
5. Total Number of Factors:

   To find the total number of factors that are multiples of 6, multiply the number of choices for each exponent:

   Total factors = 3 × 2 × 3 = 18

Answer: The number of factors of 1800 that are multiples of 6 is 18.