Explanation:

(13¹⁰⁰ + 17¹⁰⁰) = (15 – 2)¹⁰⁰ + (15 + 2)¹⁰⁰
Now 5² = 25, So, any term that has 5² or any higher power of 5 will be a multiple of 25.
So, for the above question, for computing remainder, we need to think about only the terms with 15⁰ or 15¹.
(15 – 2)¹⁰⁰ + (15 + 2)¹⁰⁰
Coefficient of 15⁰ = (-2)¹⁰⁰ + 2¹⁰⁰
Coefficient of 15¹ = ¹⁰⁰C₁ * 15¹* (-2)⁹⁹ + ¹⁰⁰C₁ * 15¹* (-2)⁹⁹.
These two terms cancel each other.So, the sum is 0.
Remainder is nothing but (-2)¹⁰⁰ + 2¹⁰⁰ = (2)¹⁰⁰ + 2¹⁰⁰
2¹⁰¹
Remainder of dividing 2¹ by 25 = 2
Remainder of dividing 2² by 25 = 4
Remainder of dividing 2³ by 25 = 8
Remainder of dividing 2⁴ by 25 = 16
Remainder of dividing 2⁵ by 25 = 32 = 7
Remainder of dividing 2¹⁰ by 25 = 7² = 49 = -1
Remainder of dividing 22⁰ by 25 = (-1)² = 1
Remainder of dividing 2¹⁰¹ by 25
= Remainder of dividing 2¹⁰⁰ by 25 * Remainder of dividing 2¹ by 25
= 1 * 2 = 2