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=9, then `np_n` equals

Options

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

Answer

To solve the problem, we need to find the value of \( n p_n \) where \( n = 10\lambda + r \) and \( r = 9 \). Let's break down the solution step by step.

### Step 1: Define \( n \)
Given that \( n = 10\lambda + r \) and \( r = 9 \), we can substitute \( r \) into the equation:
\[ n = 10\lambda + 9 \]

### Step 2: Identify the range of \( a \)
The number \( a \) is chosen from the set \( \{1, 2, 3, \ldots, n\} \). This means \( a \) can take any integer value from 1 to \( n \).

### Step 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:
\[ a^2 - 1 \equiv 0 \mod{10} \]
This implies:
\[ a^2 \equiv 1 \mod{10} \]
The values of \( a \) that satisfy this condition are those for which \( a \) ends with either 1 or 9, since:
- \( 1^2 \equiv 1 \mod{10} \)
- \( 9^2 \equiv 81 \equiv 1 \mod{10} \)

### Step 4: Count the favorable outcomes
Now we need to count how many integers \( a \) from 1 to \( n \) end with 1 or 9.

1. **Counting numbers ending with 1**:
- The numbers are \( 1, 11, 21, \ldots \) up to \( n \).
- The largest number \( \leq n \) that ends with 1 can be expressed as \( 10k + 1 \) where \( k \) is the largest integer such that \( 10k + 1 \leq n \).
- This gives \( k \leq \frac{n - 1}{10} \), so the count of such numbers is \( \left\lfloor \frac{n - 1}{10} \right\rfloor + 1 \).

2. **Counting numbers ending with 9**:
- The numbers are \( 9, 19, 29, \ldots \) up to \( n \).
- The largest number \( \leq n \) that ends with 9 can be expressed as \( 10k + 9 \) where \( k \) is the largest integer such that \( 10k + 9 \leq n \).
- This gives \( k \leq \frac{n - 9}{10} \), so the count of such numbers is \( \left\lfloor \frac{n - 9}{10} \right\rfloor + 1 \).

### Step 5: Total favorable outcomes
The total number of favorable outcomes (numbers ending with 1 or 9) is:
\[ \text{Total favorable} = \left( \left\lfloor \frac{n - 1}{10} \right\rfloor + 1 \right) + \left( \left\lfloor \frac{n - 9}{10} \right\rfloor + 1 \right) \]
This simplifies to:
\[ \text{Total favorable} = \left\lfloor \frac{n - 1}{10} \right\rfloor + \left\lfloor \frac{n - 9}{10} \right\rfloor + 2 \]

### Step 6: Calculate \( p_n \)
The total number of outcomes (total numbers from 1 to \( n \)) is \( n \). Therefore, the probability \( p_n \) is:
\[ p_n = \frac{\text{Total favorable}}{n} = \frac{\left\lfloor \frac{n - 1}{10} \right\rfloor + \left\lfloor \frac{n - 9}{10} \right\rfloor + 2}{n} \]

### Step 7: Calculate \( n p_n \)
Now we need to calculate \( n p_n \):
\[ n p_n = n \cdot \frac{\left\lfloor \frac{n - 1}{10} \right\rfloor + \left\lfloor \frac{n - 9}{10} \right\rfloor + 2}{n} = \left\lfloor \frac{n - 1}{10} \right\rfloor + \left\lfloor \frac{n - 9}{10} \right\rfloor + 2 \]

### Step 8: Substitute \( n \)
Substituting \( n = 10\lambda + 9 \) into the expression gives:
\[ n p_n = \left\lfloor \frac{10\lambda + 9 - 1}{10} \right\rfloor + \left\lfloor \frac{10\lambda + 9 - 9}{10} \right\rfloor + 2 \]
This simplifies to:
\[ n p_n = \left\lfloor \lambda + 0.8 \right\rfloor + \left\lfloor \lambda \right\rfloor + 2 = \lambda + \lambda + 2 = 2\lambda + 2 \]

### Conclusion
Thus, the final answer is:
\[ n p_n = 2\lambda + 2 \]