(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} = ^{100}C_{1} * 15^{1}* (-2)^{99} + ^{100}C_{1} * 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 = 7^{2} = 49 = -1
Remainder of dividing 22^{0} by 25 = (-1)^{2} = 1
Remainder of dividing 2^{101} by 25 = Remainder of dividing 2^{100} by 25 * Remainder of dividing 2^{1} by 25 = 1 * 2 = 2
Answer Choice (B)
Correct Answer: 2