When any two natural numbers `N_1 and N_2`, such that `N_2 = N_1 + 2` are multiplied with each other, then which digit appears least time as a unit digit if `N_2 le 1000` ?

Options

A. 0
B. 9
C. 4
D. both (a) and (c )

Answer

To solve the problem, we need to analyze the multiplication of two natural numbers \( N_1 \) and \( N_2 \) where \( N_2 = N_1 + 2 \) and \( N_2 \leq 1000 \). We want to find out which digit appears the least as a unit digit in the products of \( N_1 \) and \( N_2 \).

### Step-by-Step Solution:

1. **Define the Numbers**:
Let \( N_1 \) be any natural number. Then, \( N_2 = N_1 + 2 \).

2. **Set the Range for \( N_2 \)**:
Since \( N_2 \) must be less than or equal to 1000, we have:
\[ N_1 + 2 \leq 1000 \implies N_1 \leq 998 \]
Therefore, \( N_1 \) can take values from 1 to 998.

3. **Calculate the Product**:
The product of \( N_1 \) and \( N_2 \) is:
\[ P = N_1 \times N_2 = N_1 \times (N_1 + 2) = N_1^2 + 2N_1 \]

4. **Determine the Unit Digit**:
The unit digit of \( P \) depends on the unit digits of \( N_1 \) and \( N_2 \). We will analyze the unit digits of \( N_1 \) from 0 to 9 and calculate the corresponding unit digit of \( P \).

5. **Unit Digits of \( N_1 \)**:
- If \( N_1 \equiv 0 \mod 10 \): \( N_2 \equiv 2 \mod 10 \) → \( P \equiv 0 \times 2 \equiv 0 \mod 10 \)
- If \( N_1 \equiv 1 \mod 10 \): \( N_2 \equiv 3 \mod 10 \) → \( P \equiv 1 \times 3 \equiv 3 \mod 10 \)
- If \( N_1 \equiv 2 \mod 10 \): \( N_2 \equiv 4 \mod 10 \) → \( P \equiv 2 \times 4 \equiv 8 \mod 10 \)
- If \( N_1 \equiv 3 \mod 10 \): \( N_2 \equiv 5 \mod 10 \) → \( P \equiv 3 \times 5 \equiv 5 \mod 10 \)
- If \( N_1 \equiv 4 \mod 10 \): \( N_2 \equiv 6 \mod 10 \) → \( P \equiv 4 \times 6 \equiv 4 \mod 10 \)
- If \( N_1 \equiv 5 \mod 10 \): \( N_2 \equiv 7 \mod 10 \) → \( P \equiv 5 \times 7 \equiv 5 \mod 10 \)
- If \( N_1 \equiv 6 \mod 10 \): \( N_2 \equiv 8 \mod 10 \) → \( P \equiv 6 \times 8 \equiv 8 \mod 10 \)
- If \( N_1 \equiv 7 \mod 10 \): \( N_2 \equiv 9 \mod 10 \) → \( P \equiv 7 \times 9 \equiv 3 \mod 10 \)
- If \( N_1 \equiv 8 \mod 10 \): \( N_2 \equiv 0 \mod 10 \) → \( P \equiv 8 \times 0 \equiv 0 \mod 10 \)
- If \( N_1 \equiv 9 \mod 10 \): \( N_2 \equiv 1 \mod 10 \) → \( P \equiv 9 \times 1 \equiv 9 \mod 10 \)

6. **Collect the Unit Digits**:
From the above calculations, the possible unit digits of \( P \) are:
- 0 (from \( N_1 \equiv 0, 8 \))
- 3 (from \( N_1 \equiv 1, 7 \))
- 5 (from \( N_1 \equiv 3, 5 \))
- 4 (from \( N_1 \equiv 4 \))
- 8 (from \( N_1 \equiv 2, 6 \))
- 9 (from \( N_1 \equiv 9 \))

7. **Count the Frequency**:
- 0 appears 2 times
- 3 appears 2 times
- 5 appears 2 times
- 4 appears 1 time
- 8 appears 2 times
- 9 appears 1 time

8. **Identify the Least Frequent Digits**:
The digits that appear the least number of times (1 time) are 4 and 9.

9. **Conclusion**:
Therefore, the digit that appears the least as a unit digit when \( N_2 \leq 1000 \) is **9**.