The remainders when $2^{50}$ is divided by 7.

Step 1 (Find Cyclicity Modulo 7): We evaluate powers of 2 modulo 7:

• $2^1=2\equiv 2(\mathrm{mod}7)$

• $2^2=4\equiv 4(\mathrm{mod}7)$

• $2^3=8\equiv 1(\mathrm{mod}7)$ (This breaks the exponent into cycles of 3).

Step 2 (Exponent Division): We divide the exponent $50$ by the cycle period $3$:

Therefore:

Step 3 (Conclusion): The remainder when $2^{50}$ is divided by 7 is exactly 4.