Let ` n=10lambda+r", where " lambda,rinN, 0lerle9.` A number a is chosen at random from the set {1, 2, 3,…, n} and let `p_n` denote the probability that `(a^2-1)` is divisible by 10.
If r=0, then `np_n` equals

Options

A. `2lambda`
B. `(lambda+1)`
C. `(2lambda+1)`
D. `lambda`

Answer

To solve the problem, we need to find \( n \cdot p_n \) given that \( n = 10\lambda + r \) where \( \lambda \in \mathbb{N} \) and \( r \in \{0, 1, 2, \ldots, 9\} \). We are particularly interested in the case when \( r = 0 \).

### Step-by-Step Solution:

1. **Understanding \( n \)**:
Since \( r = 0 \), we have:
\[ n = 10\lambda + 0 = 10\lambda \]
This means \( n \) is a multiple of 10.

2. **Choosing a number \( a \)**:
A number \( a \) is chosen randomly from the set \( \{1, 2, 3, \ldots, n\} \). Thus, the total number of outcomes is \( n = 10\lambda \).

3. **Condition for divisibility**:
We need to find the probability \( p_n \) that \( a^2 - 1 \) is divisible by 10. This can be rewritten as:
\[ a^2 - 1 \equiv 0 \mod{10} \implies a^2 \equiv 1 \mod{10} \]
This means \( a^2 \) must end with the digit 1.

4. **Finding suitable values of \( a \)**:
The digits whose squares end with 1 are:
- \( 1^2 = 1 \) (ends with 1)
- \( 9^2 = 81 \) (ends with 1)

Therefore, the possible values of \( a \) that satisfy \( a^2 \equiv 1 \mod{10} \) are those that end with 1 or 9.

5. **Counting suitable values**:
In the range from 1 to \( n = 10\lambda \):
- The numbers ending with 1 are \( 1, 11, 21, \ldots, (10\lambda - 9) \) (if \( \lambda \geq 1 \)).
- The numbers ending with 9 are \( 9, 19, 29, \ldots, (10\lambda - 1) \).

Thus, there are \( \lambda \) numbers ending with 1 and \( \lambda \) numbers ending with 9. Therefore, the total favorable outcomes are:
\[ 2\lambda \]

6. **Calculating the probability \( p_n \)**:
The probability \( p_n \) is given by the ratio of favorable outcomes to total outcomes:
\[ p_n = \frac{2\lambda}{n} = \frac{2\lambda}{10\lambda} = \frac{2}{10} = \frac{1}{5} \]

7. **Finding \( n \cdot p_n \)**:
Now, we compute \( n \cdot p_n \):
\[ n \cdot p_n = 10\lambda \cdot \frac{1}{5} = 2\lambda \]

### Final Answer:
Thus, if \( r = 0 \), then \( n \cdot p_n = 2\lambda \).