Bunuel wrote:

n is a positive integer greater than 2. If y = 9^0 + 9^1 + 9^2 + … + 9^n, what is the remainder when y is divided by 5?

(1) n is divisible by 3.

(2) n is odd

\(9 \equiv -1 \pmod{5} \implies 9^k \equiv (-1)^k \pmod{5}\)

(1) \(n=3k\) so

\(\begin{split}

y &\equiv (-1)^0 &+(-1)^1 &+(-1)^2 &+...&+(-1)^{3k} &\pmod{5}\\

y &\equiv \quad 1 &+ (-1) &+\quad 1 &+...&+(-1)^k &\pmod{5}\\

\end{split}\)

If \(k\) is even, for example \(k=2\), then \(y \equiv 1 \pmod{5}\)

If \(k\) is odd, for example \(k=1\), then \(y \equiv 0 \pmod{5}\)

Insufficient.

(2) \(n=2k+1\) so

\(\begin{split}

y &\equiv (-1)^0 &+(-1)^1 &+(-1)^2 &+...&+(-1)^{2k}&+(-1)^{2k+1} &\pmod{5}\\

y &\equiv \quad 1 &+ (-1) &+\quad 1 &+...&+\quad 1 &+(-1) &\pmod{5}\\

y &\equiv \quad 0 &\pmod{5}

\end{split}\)

Sufficient

Hence the answer is B

Other solution\(y \equiv (-1)^0 +(-1)^1 +(-1)^2 +...+(-1)^{n} = \frac{(-1)^{n+1}-1}{(-1)-1}=\frac{-(-1)^{n}-1}{-2}=\frac{(-1)^n+1}{2}\pmod{5}\)

(1) If \(n=3k \implies y \equiv \frac{(-1)^{3k}+1}{2} \pmod{5}\)

If \(k=1 \implies y \equiv \frac{(-1)^3+1}{2}=0 \pmod{5}\)

If \(k=2 \implies y \equiv \frac{(-1)^6+1}{2}=1 \pmod{5}\)

Hence insufficient.

(2) If \(n=2k+1 \implies y \equiv \frac{(-1)^{2k+1}+1}{2}=\frac{(-1)+1}{2}=0 \pmod{5}\)

Hence sufficient.

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