CAT Exam

What is the last two digit of 7^2008?

Read Solution (Total 2)

7^4= 2401
7^2008= (7^4)^502= 2401^502 = (2400+01)^502

so last two digit of 7^2008 are 01.
12 years agoHelpfull: Yes(7) No(0)

7^4 ≡ 1 (mod 100)
(7^4)^502 ≡ 1 (mod 1010)

so remainder is 1
12 years agoHelpfull: Yes(1) No(5)

CAT Other Question

if 31x+30y+29z=366 then x+y+z=? where x, y, z are natural numbers. Find the probability that the sum of dots appearing in 4 successive throws of two dice is every time 13.