The remainder when 2^(2003) is divided by 17 is (A) 1 (B) 2 (C) 8 (D) ...
Solution:
We can use Fermat's Little Theorem to solve this problem.
Fermat's Little Theorem states that if p is a prime number, then for any integer a, a^p ≡ a (mod p).
In other words, if we divide a^p by p, the remainder is the same as dividing a by p.
In this problem, we have p = 17 and a = 2. Therefore, 2^17 ≡ 2 (mod 17).
We can use this result to simplify 2^2003:
2^2003 = 2^17 × 2^1986
Since 2^17 ≡ 2 (mod 17), we can write:
2^2003 ≡ 2 × (2^16)^124 ≡ 2 × 1^124 ≡ 2 (mod 17)
Therefore, the remainder when 2^2003 is divided by 17 is 2.
Answer: (B) 2
The remainder when 2^(2003) is divided by 17 is (A) 1 (B) 2 (C) 8 (D) ...
2^2003
= (2^2000)*(2^3)
= 8*(16^500)
= 8*{(17-1)^500}
= 8*{(-1)^500 + 17k}. where k is some constant
= 8 + 17*8k
now on dividing with 17 we get 8 as the reamainder
THIS QUESTION CAME IN JEE MAIN 2019 JANUARY EDITION!!