find the last two digits of the following expression 1122 1122

1122^1122!
=(2*561)^1122!
=2^1122!*561^1122!
=(2^10)^1122!/10 * 561^1122!
=(xx24)^1122!/10 * 561^1122!
we know that, 24^even = xx76
so last two digit of (xx24)^1122!/10 =76 (as 1122!/10 is even)
now last two digit of 561^1122!=01(unit digit is always 1,10th digit=10th digit of 561*unit digit of 1122!=6*0=0)
hence last two digit of 1122^1122!=76*01=76
9 years agoHelpfull: Yes(9) No(0)
1122!=1122*1121*.....*4*3*2*1..so it can be written in 4*n form.so ques reduces to 1122^4...to find last two digits we have to devide it by 100..by remainder theorem:(1122*1122*1122*1122)/100..in each case remainder will be 22..so it becomes (22*22*22*22)/100-->(484*484)/100-->(84*84)/100-->7056/100-->56..so and is 56.
10 years agoHelpfull: Yes(6) No(9)
Since power is 1122! so its remainder will be 2 when devided by 4.
So question now reduces to find last two digits of 1122^2.(i.e 1258884)
By remainder theorem,1258884/100--->84 will be remainder.so ans is 84
10 years agoHelpfull: Yes(1) No(8)
ans is 22
and correct ans...
10 years agoHelpfull: Yes(0) No(6)
4
1122! has last digit 0 and 2^0 end digit is 2
10 years agoHelpfull: Yes(0) No(6)
no of zeros in 1122!
1122/5=224+44+8+1=277 zeros
so 1122! is divisible by 4
so unit place of 1122^1122!
2^4=2
so unit place must be 2
9 years agoHelpfull: Yes(0) No(2)

find the last two digits of the following expression

(1122)^(1122!)