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!