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.

Answer 2

Step 1: Prime factorization

\[ 1800 = 18 \cdot 100 = (2 \cdot 3^2)\cdot(2^2 \cdot 5^2) = 2^3 \cdot 3^2 \cdot 5^2. \]

Step 2: General form of a divisor

Any divisor \(d\) of \(1800\) has the form \[ d = 2^{a}\,3^{b}\,5^{c},\quad \text{with } a\in\{0,1,2,3\},\; b\in\{0,1,2\},\; c\in\{0,1,2\}. \]

Step 3: Condition for being a multiple of \(6=2\cdot 3\)

For \(d\) to be divisible by \(6\), it must contain at least one factor \(2\) and one factor \(3\): \[ a \ge 1 \quad \text{and} \quad b \ge 1. \] The exponent \(c\) of \(5\) is unrestricted (can be \(0,1,2\)).

Step 4: Count admissible choices

\[ \begin{aligned} a &\in \{1,2,3\} \quad &\Rightarrow&\; 3 \text{ choices},\\ b &\in \{1,2\} \quad &\Rightarrow&\; 2 \text{ choices},\\ c &\in \{0,1,2\} \quad &\Rightarrow&\; 3 \text{ choices}. \end{aligned} \] Total number of divisors of \(1800\) that are multiples of \(6\): \[ 3 \times 2 \times 3 = 18. \]

Final Answer: \(\boxed{18}\)

Quick check (optional bijection idea)

Divisors of \(1800\) divisible by \(6\) ↔ divisors of \(\frac{1800}{6}=300=2^2\cdot 3^1\cdot 5^2\).
Number of divisors of \(300\): \((2+1)(1+1)(2+1)=3\cdot 2\cdot 3=18\) ✓