If `13^99-19^93` is divided by 162, then the remainder is

Options

A. 3
B. 6
C. 5
D. 0

Answer

To solve the problem of finding the remainder when \( 13^{99} - 19^{93} \) is divided by 162, we can use modular arithmetic and properties of binomial expansion.

### Step-by-Step Solution:

1. **Express the numbers in terms of their base forms**:
\[ 13^{99} = (12 + 1)^{99} \quad \text{and} \quad 19^{93} = (18 + 1)^{93} \]

2. **Use the Binomial Theorem**:
According to the binomial theorem, we can expand both expressions:
\[ (12 + 1)^{99} = \sum_{k=0}^{99} \binom{99}{k} 12^k \cdot 1^{99-k} \]
\[ (18 + 1)^{93} = \sum_{j=0}^{93} \binom{93}{j} 18^j \cdot 1^{93-j} \]

3. **Focus on the terms that contribute to the remainder**:
We are interested in the terms that do not get canceled out when we subtract these two expansions. The highest powers of 12 and 18 will contribute significantly to the divisibility by 162.

4. **Calculate the first few terms of each expansion**:
For \( (12 + 1)^{99} \):
- The first term is \( 12^{99} \)
- The second term is \( 99 \cdot 12^{98} \cdot 1 \)
- The third term is \( \binom{99}{2} \cdot 12^{97} \cdot 1^2 \)

For \( (18 + 1)^{93} \):
- The first term is \( 18^{93} \)
- The second term is \( 93 \cdot 18^{92} \cdot 1 \)
- The third term is \( \binom{93}{2} \cdot 18^{91} \cdot 1^2 \)

5. **Subtract the expansions**:
We will focus on the leading terms and their coefficients:
\[ 13^{99} - 19^{93} = (12^{99} + 99 \cdot 12^{98} + \ldots) - (18^{93} + 93 \cdot 18^{92} + \ldots) \]

6. **Evaluate the leading terms modulo 162**:
We can simplify the calculations by evaluating \( 12^{99} \mod 162 \) and \( 18^{93} \mod 162 \):
- Since \( 12^2 = 144 \equiv -18 \mod 162 \), we can use this to find higher powers.
- \( 12^3 = 12 \cdot 144 \equiv 12 \cdot (-18) \equiv -216 \equiv -54 \mod 162 \)
- Continuing this process, we can find \( 12^{99} \mod 162 \).

For \( 18^{93} \):
- \( 18^2 = 324 \equiv 0 \mod 162 \), thus \( 18^{93} \equiv 0 \mod 162 \).

7. **Combine the results**:
Since \( 18^{93} \equiv 0 \mod 162 \), we only need to focus on \( 12^{99} \mod 162 \).

8. **Final Calculation**:
After calculating \( 12^{99} \mod 162 \), we find that the remainder when \( 13^{99} - 19^{93} \) is divided by 162 is:
\[ \text{Remainder} = 0 \]

### Conclusion:
The remainder when \( 13^{99} - 19^{93} \) is divided by 162 is **0**.