Find a number which, when multiplied by any number of the digits, the digits in the product gives the number when these are added.

Options

A. 8
B. 7
C. 6
D. 9

Explanation

To solve the problem, we need to find a number \( F \) such that when it is multiplied by any number, the sum of the digits of the product equals \( F \). Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We need a number \( F \) such that for any integer \( n \), the sum of the digits of the product \( F \times n \) equals \( F \).

2. **Testing Small Numbers:**
We will test small integers to see if they satisfy the condition. Let's start with \( F = 1 \) and go up.

3. **Testing \( F = 1 \):**
- Multiply \( 1 \) by any number, say \( n = 5 \):
\[ 1 \times 5 = 5 \]
Sum of digits = \( 5 \) (which is equal to \( 1 \)). So, this does not satisfy.

4. **Testing \( F = 2 \):**
- Multiply \( 2 \) by \( n = 5 \):
\[ 2 \times 5 = 10 \]
Sum of digits = \( 1 + 0 = 1 \) (not equal to \( 2 \)). So, this does not satisfy.

5. **Testing \( F = 3 \):**
- Multiply \( 3 \) by \( n = 4 \):
\[ 3 \times 4 = 12 \]
Sum of digits = \( 1 + 2 = 3 \) (which is equal to \( 3 \)). This works!

6. **Testing \( F = 4 \):**
- Multiply \( 4 \) by \( n = 5 \):
\[ 4 \times 5 = 20 \]
Sum of digits = \( 2 + 0 = 2 \) (not equal to \( 4 \)). So, this does not satisfy.

7. **Testing \( F = 5 \):**
- Multiply \( 5 \) by \( n = 6 \):
\[ 5 \times 6 = 30 \]
Sum of digits = \( 3 + 0 = 3 \) (not equal to \( 5 \)). So, this does not satisfy.

8. **Testing \( F = 6 \):**
- Multiply \( 6 \) by \( n = 7 \):
\[ 6 \times 7 = 42 \]
Sum of digits = \( 4 + 2 = 6 \) (which is equal to \( 6 \)). This works!

9. **Testing \( F = 7 \):**
- Multiply \( 7 \) by \( n = 8 \):
\[ 7 \times 8 = 56 \]
Sum of digits = \( 5 + 6 = 11 \) (not equal to \( 7 \)). So, this does not satisfy.

10. **Testing \( F = 8 \):**
- Multiply \( 8 \) by \( n = 9 \):
\[ 8 \times 9 = 72 \]
Sum of digits = \( 7 + 2 = 9 \) (not equal to \( 8 \)). So, this does not satisfy.

11. **Testing \( F = 9 \):**
- Multiply \( 9 \) by \( n = 10 \):
\[ 9 \times 10 = 90 \]
Sum of digits = \( 9 + 0 = 9 \) (which is equal to \( 9 \)). This works!

### Conclusion:
The numbers \( 3, 6, \) and \( 9 \) satisfy the condition. However, since the problem asks for any number, we can conclude that \( F = 9 \) is a valid solution.