Huge number versus a modulus of 7!!!

Note first that $4671^{3928} = (667*7 + 2)^{3928} \equiv 2^{3928} \pmod{7}$ .

Next, we note that $2^{3} \equiv 1 \pmod{7}$ , and that $3928 = 1309 * 3 + 1$ .

Thus $2^{3928} = 2^{1309 * 3 + 1} = (2^{3})^{1309} * 2 \equiv 1 * 2 \pmod{7}$ .

The desired remainder is therefore $\boxed{2}$ .

4671 = 667 * 7 +2 = 2

3928 = 561 * 7 +1 = 1

So 2^1 = 2. the answer is 2

Using Fermat's Little Theorem:

$4671^{3928} \equiv 2^{3928} \equiv 2^{(7-1) \times 654+4} \equiv (2^6)^{654} \times 2^4 \equiv 1^{654} \times 16$

$\equiv 16 \equiv \boxed{2} \pmod{7}$

I jusr try 1, 2 or 3 in such type of questions. We r trained to do so in coachings to crack competitive exams