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To solve this problem, we need to determine the remainder when \( 17^{29} + 25^{29} \) is divided by 21.

We can solve this problem using modular arithmetic. Let’s break it down into steps:

### Step 1: Simplify \( 17^{29} \mod 21 \)

1. **Find \( 17 \mod 21 \)**:
   Since \( 17 < 21 \), \( 17 \mod 21 = 17 \).

2. **Use Fermat’s Little Theorem**:
   Fermat’s Little Theorem states that if \( a \) and \( p \) are coprime, then \( a^{p-1} \equiv 1 \mod p \). Here, \( p = 21 \) (not a prime), so we use Euler’s theorem instead, which involves finding \( \phi(21) = 12 \) because \( \phi(21) = \phi(3) \times \phi(7) = 2 \times 6 = 12 \).

3. **Reduce exponent**:
   Since \( 17^{12} \equiv 1 \mod 21 \), we can reduce \( 29 \mod 12 \) to get \( 29 \equiv 5 \mod 12 \).
   
   Therefore, \( 17^{29} \equiv 17^5 \mod 21 \).

4. **Calculate \( 17^5 \mod 21 \)**:
   After calculating, we find that \( 17^{29} \equiv 17 \mod 21 \).

### Step 2: Simplify \( 25^{29} \mod 21 \)

Following a similar approach:
1. **Find \( 25 \mod 21 = 4 \)**.

2.

What is the remainder, if (17^29 + 25^29) is divided by 21?

Explore how to find the remainder of (17^29 + 25^29) when divided by 21 using modular arithmetic, Euler's theorem, and Fermat's Little Theorem. Suitable for students in Grades 10-12, this problem illustrates key concepts in number theory and exponentiation.

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