Let `n = 2^3 4^5 6^8 5^4`. A positive factor is taken atrandom from the possible positive factors of n.Then the probability that the selected factor is aperfect square and divisible by 100 is

Options

A. `5/99`
B. `10/99`
C. `1/6`
D. `71/99`

Answer

To solve the problem, we need to find the probability that a randomly selected positive factor of \( n = 2^3 \cdot 4^5 \cdot 6^8 \cdot 5^4 \) is a perfect square and divisible by 100.

### Step 1: Factorization of \( n \)

First, we will express \( n \) in terms of its prime factors:

\[ n = 2^3 \cdot 4^5 \cdot 6^8 \cdot 5^4 \]

We can rewrite \( 4 \) and \( 6 \) in terms of their prime factors:

\[ 4 = 2^2 \quad \text{and} \quad 6 = 2 \cdot 3 \]

Now substituting these into the expression for \( n \):

\[ n = 2^3 \cdot (2^2)^5 \cdot (2 \cdot 3)^8 \cdot 5^4 \]

This simplifies to:

\[ n = 2^3 \cdot 2^{10} \cdot 2^8 \cdot 3^8 \cdot 5^4 \]

Now, we can combine the powers of \( 2 \):

\[ n = 2^{3 + 10 + 8} \cdot 3^8 \cdot 5^4 = 2^{21} \cdot 3^8 \cdot 5^4 \]

### Step 2: Total Number of Factors of \( n \)

The total number of positive factors of \( n \) can be calculated using the formula:

\[ (\text{exponent of } p_1 + 1)(\text{exponent of } p_2 + 1)(\text{exponent of } p_3 + 1) \]

For \( n = 2^{21} \cdot 3^8 \cdot 5^4 \):

\[ \text{Total factors} = (21 + 1)(8 + 1)(4 + 1) = 22 \cdot 9 \cdot 5 = 990 \]

### Step 3: Conditions for Perfect Square and Divisible by 100

A factor is a perfect square if all the exponents in its prime factorization are even.

To be divisible by \( 100 \), a factor must include at least \( 2^2 \) and \( 5^2 \).

Thus, we can express a perfect square factor of \( n \) that is divisible by \( 100 \) as:

\[ 2^{2a} \cdot 3^{2b} \cdot 5^{2c} \]

Where:
- \( 2a \geq 2 \) (so \( a \geq 1 \))
- \( 2c \geq 2 \) (so \( c \geq 1 \))
- \( 2a \leq 21 \) (so \( a \leq 10 \))
- \( 2b \leq 8 \) (so \( b \leq 4 \))
- \( 2c \leq 4 \) (so \( c \leq 2 \))

### Step 4: Counting Valid Values for \( a, b, c \)

- For \( a \): Possible values are \( 1, 2, \ldots, 10 \) (10 options)
- For \( b \): Possible values are \( 0, 1, 2, 3, 4 \) (5 options)
- For \( c \): Possible values are \( 1, 2 \) (2 options)

Thus, the total number of favorable outcomes is:

\[ 10 \cdot 5 \cdot 2 = 100 \]

### Step 5: Probability Calculation

Now, we can calculate the probability that a randomly selected factor is a perfect square and divisible by \( 100 \):

\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of factors}} = \frac{100}{990} = \frac{10}{99} \]

### Final Answer

The probability that the selected factor is a perfect square and divisible by \( 100 \) is:

\[ \frac{10}{99} \]