A number N is formed by writing 9 for 99 times. What is the remainder if N is divided by 13 ?

• The number N consists of the digit 9 repeated 99 times, which can be expressed as $9 \times \frac{10^{99}-1}{9} = 10^{99}-1$.

• To find the remainder of $10^{99}-1$ when divided by 13, we first determine the remainder of $10^{99}$ modulo 13.

• Observe the powers of 10 modulo 13: $10^1 \equiv 10$, $10^2 \equiv 100 \equiv 9$, $10^3 \equiv 90 \equiv 12 \equiv -1$, and $10^6 \equiv (-1)^2 \equiv 1$.

• Since $10^6 \equiv 1 \pmod{13}$, we can reduce the exponent 99 by dividing by 6: $99 = 6 \times 16 + 3$.

• Thus, $10^{99} \equiv (10^6)^{16} \times 10^3 \equiv 1^{16} \times (-1) \equiv -1 \pmod{13}$.

• Therefore, $N = 10^{99}-1 \equiv -1 - 1 \equiv -2 \equiv 11 \pmod{13}$, so the remainder is 11.