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The expression 2222^(7777) +7777^(2222) ...

The expression `2222^(7777) +7777^(2222)` is divisible by :

A

99

B

101

C

13

D

any two of these

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AI Generated Solution

The correct Answer is:
To determine the divisibility of the expression \( 2222^{7777} + 7777^{2222} \), we will analyze the numbers involved and check their divisibility by the given options. ### Step-by-Step Solution: 1. **Identify the Numbers**: - We have two numbers: \( 2222 \) and \( 7777 \). 2. **Check Divisibility by 101**: - First, we note that \( 2222 \) can be expressed as \( 22 \times 101 \) (since \( 2222 = 2 \times 1000 + 2 \times 200 + 2 \times 20 + 2 \times 2 \)). - Similarly, \( 7777 \) can be expressed as \( 77 \times 101 \). - Since both \( 2222 \) and \( 7777 \) are multiples of \( 101 \), any power of these numbers will also be divisible by \( 101 \). - Therefore, \( 2222^{7777} \) is divisible by \( 101 \) and \( 7777^{2222} \) is also divisible by \( 101 \). - Thus, their sum \( 2222^{7777} + 7777^{2222} \) is divisible by \( 101 \). 3. **Check Divisibility by 99**: - A number is divisible by \( 99 \) if it is divisible by both \( 9 \) and \( 11 \). - **Divisibility by 11**: - For \( 2222 \), the sum of the digits is \( 2 + 2 + 2 + 2 = 8 \) (which is not divisible by \( 11 \)). - For \( 7777 \), the sum of the digits is \( 7 + 7 + 7 + 7 = 28 \) (which is also not divisible by \( 11 \)). - Therefore, \( 2222 \) and \( 7777 \) are not divisible by \( 11 \). - Since neither number is divisible by \( 11 \), their sum cannot be divisible by \( 99 \). 4. **Check Divisibility by 13**: - We can check the divisibility of \( 2222 \) and \( 7777 \) by \( 13 \). - \( 2222 \div 13 \approx 171.692 \) (not divisible). - \( 7777 \div 13 \approx 599.0 \) (divisible). - Since \( 2222 \) is not divisible by \( 13 \), their sum cannot be divisible by \( 13 \). 5. **Conclusion**: - The expression \( 2222^{7777} + 7777^{2222} \) is divisible by \( 101 \) but not by \( 99 \) or \( 13 \). - Therefore, the correct answer is that the expression is divisible by \( 101 \). ### Final Answer: The expression \( 2222^{7777} + 7777^{2222} \) is divisible by \( 101 \).
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