What is the remainder when 44^(197) divided by 7?

To find the remainder when $44^{197}$ is divided by 7, we can use modular arithmetic and properties of congruences.

Step 1: Simplify the base modulo 7. $44 \equiv 44 \mod 7$ $44 = 7 \times 6 + 2$ $44 \equiv 2 \mod 7$

So, $44^{197} \equiv 2^{197} \mod 7$.

Step 2: Use Fermat's Little Theorem, which states that if $p$ is a prime number and $a$ is an integer not divisible by $p$, then $a^{p-1} \equiv 1 \mod p$.

Here, $p = 7$ and $a = 2$. According to Fermat's Little Theorem: $2^{6} \equiv 1 \mod 7$

Step 3: Express 197 in terms of multiples of 6. $197 = 6 \times 32 + 5$

So, $2^{197} = 2^{6 \times 32 + 5} = (2^6)^{32} \times 2^5$

Using Fermat's Little Theorem: $(2^6)^{32} \equiv 1^{32} \equiv 1 \mod 7$

Therefore, $2^{197} \equiv 1 \times 2^5 \mod 7$ $2^{197} \equiv 2^5 \mod 7$

Step 4: Calculate $2^5 \mod 7$. $2^5 = 32$ $32 \equiv 32 \mod 7$ $32 = 7 \times 4 + 4$ $32 \equiv 4 \mod 7$

So, the remainder when $44^{197}$ is divided by 7 is $\boxed{4}$.