The Greatest 91

Let $x=\left(\sqrt{91}+1\right)^{91}-\left(\sqrt{91}-1\right)^{91}$ . We want to find $x \mod 10$ . We have: $\begin{aligned} x &\equiv \sum_{k=0}^{91} \binom{91}{k} \sqrt{91}^k-\sum_{k=0}^{91} \binom{91}{k} \sqrt{91}^k(-1)^{k+1} \pmod {10}\\ &\equiv 2\sum_{k=0}^{45} \binom{91}{2k} \sqrt{91}^{2k} \pmod {10}\\ &\equiv 2\sum_{k=0}^{45} \binom{91}{2k} {91}^k \pmod {10}\\ &\equiv 2\sum_{k=0}^{45} \binom{91}{2k} \pmod {10}\\ \end{aligned}$ But we have $\displaystyle (1+1)^{91}=\sum_{k=0}^{91} \binom{91}{k}$ and $\displaystyle (1-1)^{91}=\sum_{k=0}^{91} \binom{91}{k}(-1)^{k}$ , also $2^5 \equiv 32 \equiv 2 \pmod {10}$ , so: $\begin{aligned} x &\equiv (1+1)^{91}+(1-1)^{91} \pmod {10}\\ &\equiv 2^{91} \pmod{10}\\ &\equiv 2^{5 \cdot 18+1} \pmod{10}\\ &\equiv 2^{19} \pmod{10}\\ &\equiv 2^{5 \cdot 3+4} \pmod{10}\\ &\equiv 2^7 \pmod{10}\\ &\equiv 2^{5 \cdot 1+2} \pmod{10}\\ &\equiv 2^3 \pmod{10}\\ &\equiv \boxed{8} \pmod{10} \end{aligned}$

First, one must recognise that the terms of the second part expanded are just the same as the terms in the first part expanded multiplied by either -1 or 1 (in alternation), one must also realise that the terms divided by the binomial coefficients in both sequence alternate between powers of 91 and irrational powers of the square root of 91. These two facts mean that all irrational terms divided by the binomial coefficients are positive in the second part (so are subtracted, so cancel out with the first part), and all the powers of 91 are negative (so are subtracted to the terms in the first part, meaning all the terms in the first part are multiplied by 2). We must than realise that all powers of 91 will end in 1 (as the effect of all digits besides the first digit in the sum equivalent of the product with another integer (91 in this case) will always be a multiple of 10), so we can treat all of these terms divided by the coefficients as effectively 1 (as the effect of all digits besides the first digit in the sum equivalent of the product with another integer (the coefficients in this case) will always be a multiple of 10) since we only care about the last digit. You must than realise that there are 92 terms in first part, and the first 46 coefficients are the same as the next 46 coefficients, and the set of all the coefficients cancelled in the second 46 coefficients is the same as the set of all those coefficients not cancelled in the first 46 coefficients (because of the nature of the alternations and symmetry), and since we are multiplying the non cancelled terms by 2, and treating the powers of 91 as 1, all we are left with are all the possible binomial coefficients when raising by 91, which is the same as if both terms that are being raised by 91 are 1, which implies we are dealing with 2^91. 2^5=32, and since the last digit is 2, from the same logic that we used for powers of 91, the pattern is repeated, so powers of two repeat them selves after every 5-1=4 powers of two, and since 88/4=22 which is an integer, and 91=88+3, we can treat our sum as 2^3=8.