A number lock can be unlocked by a 3 digit code, each digit of the code could be any one of 0, 1 ....,9. What is the maximum number of trials required to unlock, if the following are known (a) the last digit (b) the sum of the first two digits is always less than or equal to the third?

Options

A. A) 9
B. B) 45
C. C) 55
D. D) 36

Answer

To solve the problem, we need to determine the maximum number of trials required to unlock a 3-digit code given the conditions that the last digit is known and the sum of the first two digits is always less than or equal to the third digit.

### Step-by-step Solution:

1. **Understanding the Code Structure**:
- The code consists of three digits: \( a \), \( b \), and \( c \).
- Each digit can take any value from 0 to 9.
- The last digit \( c \) is known, and we need to find valid combinations of \( a \) and \( b \) such that \( a + b \leq c \).

2. **Identifying the Range for \( c \)**:
- Since \( c \) can be any digit from 0 to 9, we will analyze each possible value of \( c \) from 0 to 9.

3. **Calculating Valid Combinations**:
- For each value of \( c \), we need to find how many pairs \( (a, b) \) satisfy the condition \( a + b \leq c \).
- The number of valid combinations for each \( c \) can be calculated as follows:
- If \( c = 0 \): The only valid pair is \( (0, 0) \) → 1 combination.
- If \( c = 1 \): Valid pairs are \( (0, 0) \) and \( (0, 1), (1, 0) \) → 2 combinations.
- If \( c = 2 \): Valid pairs are \( (0, 0), (0, 1), (1, 0), (0, 2), (2, 0), (1, 1) \) → 3 combinations.
- Continuing this way, we find:
- \( c = 3 \): 4 combinations
- \( c = 4 \): 5 combinations
- \( c = 5 \): 6 combinations
- \( c = 6 \): 7 combinations
- \( c = 7 \): 8 combinations
- \( c = 8 \): 9 combinations
- \( c = 9 \): 10 combinations

4. **Summing Up the Combinations**:
- Now, we sum the valid combinations for each value of \( c \):
\[ \text{Total combinations} = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 \]
- This is the sum of the first 10 natural numbers, which can be calculated using the formula:
\[ S_n = \frac{n(n + 1)}{2} \]
- Here, \( n = 10 \):
\[ S_{10} = \frac{10 \times 11}{2} = 55 \]

5. **Conclusion**:
- Therefore, the maximum number of trials required to unlock the lock is **55**.