find last two digits of 1122 1122

(1122)^(1122!)=(2*561)^1122!=(2^10)^(1122!/10) * (561)^1122!
last 2 digit of (2^10)^(1122!/10)= (XX24)^even no= 76
last 2 digit (561)^1122!=01

therefore last digit =76*01= 76
ans 76
10 years agoHelpfull: Yes(10) No(4)
last 2 digits of 1122! is 00.
2^10=1024
24^even=76
so ans is 76
10 years agoHelpfull: Yes(3) No(2)
(11*2)1122 then (2^10)112*2^2) which is 4 and 11^1122 is 21 so 21*4=84
10 years agoHelpfull: Yes(1) No(5)
76 ans is
10 years agoHelpfull: Yes(1) No(3)
Can u explain dis plz
10 years agoHelpfull: Yes(1) No(1)
1122! : last two digits = 00.
10 years agoHelpfull: Yes(1) No(5)
as it is 1122! in the power.. it will be 01
10 years agoHelpfull: Yes(1) No(2)
22^2=xx84.... power cycle of 84 will repeat after each 10 term... thus ans is 83.(bcz 1122! contains so many zeros i.e multiples of 10)
10 years agoHelpfull: Yes(1) No(0)
ans is 76
10 years agoHelpfull: Yes(0) No(0)
76 is the answer.....
10 years agoHelpfull: Yes(0) No(0)

find last two digits of 1122^1122! ??