We are asked if $3^{444} + 4^{333}$ is divisible by 5. To check this, we can use modular arithmetic.

Step-by-Step Solution:

We want to calculate $3^{444} + 4^{333} \mod 5$, i.e., the remainder when $3^{444} + 4^{333}$ is divided by 5.

Step 1: Find $3^{444} \mod 5$

Using Fermat's Little Theorem, we know that if $a$ is not divisible by $p$, then: $a^{p-1} \equiv 1 \mod p$ For $p = 5$, this tells us $3^4 \equiv 1 \mod 5$.

So, $3^{444} = (3^4)^{111} \times 3^0 \equiv 1^{111} \times 1 \equiv 1 \mod 5$.

Step 2: Find $4^{333} \mod 5$

Similarly, we can use Fermat’s Little Theorem for $4$ as well. Since $4^4 \equiv 1 \mod 5$, we know: $4^{333} = 4^{4 \times 83 + 1} = (4^4)^{83} \times 4^1 \equiv 1^{83} \times 4 \equiv 4 \mod 5$

Step 3: Add the results modulo 5

Now we compute the sum modulo 5: $3^{444} + 4^{333} \equiv 1 + 4 \equiv 5 \equiv 0 \mod 5$

Conclusion:

Yes, $3^{444} + 4^{333}$ is divisible by 5.

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Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. How does Fermat's Little Theorem help in modular arithmetic?
2. Can we generalize this method for any prime divisor?
3. What happens if we try dividing by numbers other than 5?
4. How would the solution change if the exponents were different?
5. Can this method be used for larger bases and powers efficiently?

Tip: Modular arithmetic often simplifies large exponentiation problems by reducing the problem into manageable cycles or periodic patterns.