$51 \equiv 2 \pmod 7$

$2^3 = 8 \equiv 1 \pmod 7$

$51^{203} \equiv 2^{203} = (2^3)^{67} \times 2^2 \equiv 1^{67} \times 4 = 4 \pmod 7$

Correct Option: (a) 4

$59 \equiv 3 \pmod 7$

$3^3 = 27 \equiv -1 \pmod 7$

$59^{28} \equiv 3^{28} = (3^3)^9 \times 3 \equiv (-1)^9 \times 3 = -3 \equiv 4 \pmod 7$

Correct Option: (b) 4

$67 \equiv 4 \equiv -3 \pmod 7$

$67^{99} \equiv (-3)^{99} = -(3^3)^{33} \equiv -(-1)^{33} = 1 \pmod 7$

Correct Option: (d) 1

$75 \equiv 5 \equiv -2 \pmod 7$

$75^{80} \equiv (-2)^{80} = 2^{80} = (2^3)^{26} \times 2^2 \equiv 1^{26} \times 4 = 4 \pmod 7$

Correct Option: (a) 4

$41 \equiv -1 \pmod 7$

$41^{77} \equiv (-1)^{77} = -1 \equiv 6 \pmod 7$

Correct Option: (c) 6

$21 \equiv 4 \pmod{17}$

$21^{875} \equiv 4^{875} = 2^{1750} \pmod{17}$

By Fermat's Little Theorem, $2^{16} \equiv 1 \pmod{17}$.

$1750 \pmod{16} = 6 \implies 2^{1750} \equiv 2^6 = 64 \equiv 13 \pmod{17}$

Correct Option: (b) 13

$54 \equiv 3 \pmod{17}$

$3^4 = 81 \equiv -1 \pmod{17}$

$54^{124} \equiv 3^{124} = (3^4)^{31} \equiv (-1)^{31} = -1 \equiv 16 \pmod{17}$

$83 \equiv -2 \pmod{17}$

$83^{261} \equiv (-2)^{261} = -2^{261} \pmod{17}$

By Fermat's Little Theorem, $2^{16} \equiv 1 \pmod{17}$.

$261 \pmod{16} = 5 \implies 2^{261} \equiv 2^5 = 32 \equiv 15 \pmod{17}$

Remainder $= -15 \equiv 2 \pmod{17}$

Correct Option: (d) 2

$25 \equiv 8 \pmod{17}$

$25^{102} \equiv 8^{102} = (2^3)^{102} = 2^{306} \pmod{17}$

By Fermat's Little Theorem, $2^{16} \equiv 1 \pmod{17}$.

$306 \pmod{16} = 2 \implies 2^{306} \equiv 2^2 = 4 \pmod{17}$

Correct Option: (c) 4

To find last two digits, find remainder modulo $100$:

Divide numerator and denominator by $25$: denominator becomes $4$, $125$ becomes $5$.

Divide $122$ and denominator $4$ by $2$: denominator becomes $2$, $122$ becomes $61$.

Now evaluate modulo $2$:

Multiply back by total simplified factor ($25 \times 2 = 50$):

Correct Option: (b) 50