To solve \( 5^{40} \mod 49 \) (the remainder when \( 5^{40} \) is divided by 49), we can use modular arithmetic and a method called **Euler's Theorem** or break the problem down into smaller parts to simplify the exponentiation.

### Step-by-step solution:

1. **Euler’s Theorem** tells us that if \( a \) and \( n \) are coprime (i.e., \( \gcd(a, n) = 1 \)), then:
   \[
   a^{\phi(n)} \equiv 1 \ (\text{mod}\ n)
   \]
   where \( \phi(n) \) is Euler's totient function of \( n \), which gives the number of integers less than \( n \) that are coprime with \( n \).

2. First, calculate \( \phi(49) \). Since \( 49 = 7^2 \), and for a prime \( p \):
   \[
   \phi(p^k) = p^k - p^{k-1}
   \]
   Thus,
   \[
   \phi(49) = 49 - 7 = 42
   \]
   So, by Euler’s theorem:
   \[
   5^{42} \equiv 1 \ (\text{mod}\ 49)
   \]
   This simplifies the original exponent, because we only need to calculate \( 5^{40} \mod 49 \).

3. **Simplifying the exponent**:
   \[
   5^{40} = 5^{42} \cdot 5^{-2}
   \]
   Since \( 5^{42} \equiv 1 \ (\text{mod}\ 49) \), we just need to compute:
   \[
   5^{-2} \mod 49
   \]
   Finding the modular inverse of \( 5^2 = 25 \mod 49 \).

Let me calculate this result for you.The remainder when \( 5^{40} \) is divided by 49 is \( \boxed{2} \).

Would you like more details on the process or any further clarifications?

Here are 5 related questions to explore further:
1. What is Euler's Theorem and how does it simplify modular arithmetic?
2. How do you compute the totient function \( \phi(n) \) for any number \( n \)?
3. What are modular inverses, and how are they useful in problems like this?
4. How does Fermat's Little Theorem relate to solving modular exponentiation?
5. Can you calculate \( 5^{100} \mod 49 \) using the same methods?

**Tip**: Modular exponentiation can be made much faster using techniques like **exponentiation by squaring**!

Solve the modular arithmetic problem 5^40 mod 49 step-by-step using Euler's Theorem and modular inversion. Learn key number theory concepts and simplify exponentiation using modular techniques. Suitable for Grades 10-12.

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