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

Question

Accepted Answer

The correct answer is option A. The number N consists of 99 nines. This can be written as N = 9 * (10^99 - 1) / 9 = 10^99 - 1. We need to find (10^99 - 1) mod 13.
First, let's find the cycle of powers of 10 modulo 13:
10^1 ≡ 10 (mod 13)
10^2 ≡ 100 ≡ 9 (mod 13)
10^3 ≡ 10 * 9 = 90 ≡ 12 (mod 13) or -1 (mod 13)
10^4 ≡ 10 * 12 = 120 ≡ 3 (mod 13)
10^5 ≡ 10 * 3 = 30 ≡ 4 (mod 13)
10^6 ≡ 10 * 4 = 40 ≡ 1 (mod 13)
The cycle length (order) of 10 modulo 13 is 6.
Now, divide the exponent 99 by the cycle length 6: 99 = 6 × 16 + 3.
So, 10^99 ≡ (10^6)^16 × 10^3 ≡ 1^16 × 10^3 ≡ 10^3 (mod 13).
From our cycle, 10^3 ≡ 12 (mod 13).
Therefore, N ≡ (10^99 - 1) ≡ (12 - 1) ≡ 11 (mod 13).
 The remainder is 11.

Q17. A number N is formed by writing 9 for 99 times. What is the remainder if N is divided by 13 ?

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