Correct option is (b) 8
To elaborate: Consider the binomial expansion of (m+1)^71 and (m-1)^71 which gives these two
expressions below respectively: 1) m^51 + ^51C1m^50 + ^51C2m^49 + ^51C3m^48 + … + ^51C50m^1 + ^51C51m^0
2) m^51 – ^51C1m^50 + ^51C2m^49 – ^51C3m^48 + … + ^51C50m^1 – ^51C51m^0 .
By taking the difference we have, 2(^51C1m^50 + ^51C3m^48 – ^51C5m^46 + … + ^51C50m^2 – ^51C51m^0 ).
In this case, m = \(\sqrt{51}\) and 2(^51C1m^50 + ^51C3m^48 – ^51C5m^46 + … + ^51C50m^2 – ^51C51m^0 ).
Consider, module 10 on the powers(for any natural number n): (51)^n ≡ (51 mod 10^n) ≡ 1 gives 2(^51C1 + ^51C3 + ^51C5 + … + ^51C50 + ^51C51). Now, by adding the odd terms of the 51^st row of the Pascal Triangle 2.(\(\frac{1}{2}\) * 2^51) = 2^51 = 2^(51 mod 4) = 2^3 = 8.