1 Crore+ students have signed up on EduRev. Have you? 
(13^{100} + 17^{100}) = (15 – 2)^{100} + (15 + 2)^{100}
Now 5^{2 }= 25, So, any term that has 5^{2 }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 15^{0} or 15^{1}.
(15 – 2)^{100} + (15 + 2)^{100}
Coefficient of 15^{0 }= (2)^{100} + 2^{100}
Coefficient of 15^{1} = 100C1 * 15^{1}* (2)^{99} + 100C1 * 15^{1}* (2)^{99}.
These two terms cancel each other. So, the sum is 0.
Remainder is nothing but (2)^{100} + 2^{100} = (2)^{100} + 2^{100}
2^{101}
Remainder of dividing 2^{1} by 25 = 2
Remainder of dividing 2^{2 }by 25 = 4
Remainder of dividing 2^{3} by 25 = 8
Remainder of dividing 2^{4} by 25 = 16
Remainder of dividing 2^{5} by 25 = 32 = 7
Remainder of dividing 2^{10} by 25 = 72 = 49 = 1
Remainder of dividing 2^{20} by 25 = (1)^{2} = 1
Remainder of dividing 2^{101} by 25
= Remainder of dividing 2^{100} by 25 * Remainder of dividing 21 by 25
= 1 * 2 = 2
Hence the answer is "2".
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 