Find the remainder? when 2^2011 is divisible by 13
To find the remainder when \(2^{2011}\) is divided by 13, we can use the concept of modular arithmetic.
Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" when they reach a certain value. In other words, when we divide a number by a certain modulus (in this case, 13), we are interested in the remainder.
To find the remainder when \(2^{2011}\) is divided by 13, we can simplify it by using the properties of modular arithmetic:
1. Start with the base, 2.
2. Compute the powers of 2 modulo 13: \(2^1 \equiv 2 \pmod{13}\), \(2^2 \equiv 4 \pmod{13}\), \(2^3 \equiv 8 \pmod{13}\), \(2^4 \equiv 3 \pmod{13}\), \(2^5 \equiv 6 \pmod{13}\), and \(2^6 \equiv 12 \pmod{13}\).
3. Notice that the powers of 2 repeat with a period of 6: \(2^7 \equiv 1 \pmod{13}\), \(2^8 \equiv 2 \pmod{13}\), and so on.
4. Since 2011 is not a multiple of 6, we can use the fact that \(2^7 \equiv 1 \pmod{13}\) to simplify the expression: \(2^{2011} \equiv 2^{6 \times 335 + 1} \equiv (2^6)^{335} \times 2^1 \equiv 1^{335} \times 2 \equiv 2 \pmod{13}\).
Therefore, the remainder when \(2^{2011}\) is divided by 13 is 2.