Modular Arithmetic Question 2

$\large{\text{Last digit of } 9^{87} = 9^{87\pmod{2}} =9^1=9}$ $\large{\text{Last digit of } 9^{87} +1 \equiv 9+1 \pmod{10} =0}$ $\large{\text{Last digit of 0}\implies \text{Divisible by 10}}$

Powers of $9$ follow a particular pattern. If the power be odd, then the unit's place digit of the result is $9$ . If the power be even, then the unit's place digit of the result is $1$ .

Since $87$ is odd, the unit's place digit of $9^{87}$ is $9$ , and hence the unit's place digit of $9^{87}+1$ is $0$ .

So the number is divisible by $10$ , and the correct answer is Yes .

$\large 9 ≡ -1 \hspace{0.5cm} \text{(mod 10)}$ $\large x = 9^{87} + 1 ≡ (-1)^{87} + 1 ≡ -1 + 1 ≡ 0 \hspace{0.5cm} \text{(mod 10)}$ $\large \text{Therefore, it is divisible by 10.}$

x = (10-1)^87 + 1. Expanding quantity in parenthesis via the binomial theorem and factoring out a 10 from each term that has one, we find:

x = 10*(some integer) + (-1)^87 + 1.

Since 87 is odd, (-1)^87 = -1 so that x is divisible by 10.