Let there be a random number n such that the product of its digits is 6. How many values of n are there, if n lies between `10^9` and `10^10`?

Options

A. A) 10
B. B) 60
C. C) 90
D. D) 100

Answer

To solve the problem of finding how many random numbers \( n \) exist such that the product of its digits is 6 and \( n \) lies between \( 10^9 \) and \( 10^{10} \), we can break it down into systematic steps.

### Step 1: Understand the constraints
Since \( n \) lies between \( 10^9 \) and \( 10^{10} \), it must be a 10-digit number. This means that \( n \) can be represented as \( d_1 d_2 d_3 d_4 d_5 d_6 d_7 d_8 d_9 d_{10} \), where each \( d_i \) is a digit from 0 to 9.

### Step 2: Identify the possible digit combinations
The product of the digits must equal 6. We can express 6 as a product of digits in the following ways:
1. \( 6 = 6 \)
2. \( 6 = 2 \times 3 \)
3. \( 6 = 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 6 \) (where we have nine 1's and one 6)

### Step 3: Case 1 - Using the digit 6
In this case, we use one digit as 6 and the remaining nine digits as 1's:
- The number can be represented as \( 1, 1, 1, 1, 1, 1, 1, 1, 1, 6 \).
- The digit 6 can occupy any of the 10 positions.

**Calculating combinations:**
- The number of ways to arrange this is simply the number of positions for the digit 6, which is 10.

### Step 4: Case 2 - Using the digits 2 and 3
In this case, we use the digits 2 and 3, and the remaining eight digits will be 1's:
- The number can be represented as \( 1, 1, 1, 1, 1, 1, 1, 2, 3, 1 \).
- We need to arrange the digits 2 and 3 among the 10 positions.

**Calculating combinations:**
- The number of ways to choose 2 positions out of 10 for the digits 2 and 3 can be calculated using combinations:
\[ \text{Number of arrangements} = \binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \]

### Step 5: Total combinations
Now, we add the combinations from both cases:
- From Case 1: 10
- From Case 2: 45

So, the total number of valid 10-digit numbers \( n \) such that the product of its digits is 6 is:
\[ 10 + 45 = 55 \]

### Final Answer
Thus, the total number of values of \( n \) is **55**.

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