Question

Compute the last two digits to the right of $7^{111}}$ (probably have to use modular arithmetic)

Answer 1

Here we have to find last two digits on the right of 7^111. They are given by Last two digits of any number is given by when we divide that number by 100 and the remainder is the last two digits of that number. For this we have to find 7^111 (mod 100) Here gcd (7, 100) = 1 as 7 is prime & 7 times 100. Therefore By Eulers theorem, 7^phi(100) = 1 (mod 100) 100 = 2^2 times 5^2 phi(100) = 100 (1 - 1/2)(1 - 1/5) = 4 times 10 = 40 therefore 7^40 = 1 (mod 100) therefore 7^80 = 1 (mod 100) 7^3 = 43 (mod 100) 7^6 = 49 (mod 100) (because 7^3 times 7^3 = (43 times 43) (mod 100)) 43 times 43 = 49 (mod 100) 7^6 times 7^6 = (49 times 49) (mod 100) 49 times 49 = 2401 and 2401 = 1 (mod 100)