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(#M40025970) TCS Logical Reasoning question Keep an EYE

What will be the last two digit of 1122^1122!

Asked In TCS TargetTCS 2014 (11 years ago)

Unsolved Read Solution (16)

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Read Solution (16) :

Kundan Kumar  (11 years ago)

1122! have more than 2 two's hence it is a multiple of 4..
now just calculate the last two digits of 1122^4
it will be 56( by remainder theorem)
since 1122! has ...000 at the end . Therefore it is a multiple of 10
So,
56^1= ...56
56^2= ...36
similarly,
56^5=....76(last two digit only)
now, 56^10=...76(last two digit only)
Hence last two digit of 1122^1122! will be 76.

Is this solution Helpfull? Yes (20) | No (20)  

Smuti ranjan mahapatra  (11 years ago)

(2^10)^even=76
1121! is a even no

Is this solution Helpfull? Yes (11) | No (9)  

abhinav  (11 years ago)

answer is 84
1122^1122== (102x11)^1122
--> (((102^10)^112)x2^2 x 11^1122)
-->(24^even x 2^2 x 21)
---> 76x04x21-----> 84.......awesome

Is this solution Helpfull? Yes (10) | No (21)  

D G  (11 years ago)

1122^40*28+2....so last 2 digit=(1122)^2= ans is 84

Is this solution Helpfull? Yes (9) | No (21)  

Kundan Kumar  (11 years ago)

84 cannot be the answer.
because if we calculate the unit digit of the same problem then it is coming 6
84 is the wrong answer..

Is this solution Helpfull? Yes (9) | No (9)  

Raushan Kumar  (11 years ago)

(1122)^1122!=(22)^1122!=(11)^1122! * (2)^1122!
11^1122!=01
2^1122!=76
So,76*01=76

Is this solution Helpfull? Yes (6) | No (10)  

Ankit Gorakhpuriya  (10 years ago)

If there is a base ending with 2, then each 20th power will end with 76 as last two digits. so upto 1122^1120, last two digits would be 76. and for 1122^1122,
........76 * 1122 gives .........72
........72 * 1122 gives .........84.
So answer is 84.

Is this solution Helpfull? Yes (4) | No (4)  

Chandra Bhatt  (11 years ago)

infinity ! :D

Is this solution Helpfull? Yes (3) | No (15)  

Rincy juliet  (11 years ago)

1122! have more than 2 two's hence it is a multiple of 4..
now just calculate the last two digits of 1122^4
it will be 56( by remainder theorem)
since 1122! has ...000 at the end . Therefore it is a multiple of 10
So,
56^1= ...56
56^2= ...36
similarly,
56^5=....76(last two digit only)
now, 56^10=...76(last two digit only)
Hence last two digit of 1122^1122! will be 76 :)

Is this solution Helpfull? Yes (3) | No (4)  

SHIVAM  (11 years ago)

4 multiply it

Is this solution Helpfull? Yes (2) | No (4)  

sailakshmi  (10 years ago)

1122^1122
==1122 can be written as 102*11
==(102*11)^1122
==(102^1122)*(11^1122) /*11 power anynuber results 21 as last two digits*/
==(102^(1120+2))*(21)
==(102^10^112)*(102^2)*(21)
==(24^112)(04)(21) /* here 102^2 gives 04 as last two digits and 102^10 gives 24 as last two digits 2^10 is 1024*/
==76*04*21 /* here 24 ^even is always 76*/
==....84

Is this solution Helpfull? Yes (1) | No (5)  

upasana acharya  (10 years ago)

44
2*2 lst digit
2nd last 2*2

Is this solution Helpfull? Yes (1) | No (1)  

Akhil Sharmaq  (11 years ago)

4 is the answer

Is this solution Helpfull? Yes | No (7)  

RAJA RAVI THEJA KALLA  (10 years ago)

according to arun sharma book if 1122! is at power then divide it with a and we get remainder of 2 and last digit is 2 in unit place of base so unit digit is of 4

Is this solution Helpfull? Yes | No (3)  

subhasree  (10 years ago)

unit place 2^2=4;tens digit 2*2=4
so ans. is 44

Is this solution Helpfull? Yes | No (2)  

ANKIT KOTNALA  (10 years ago)

Answer will be "04"

See it has a concept as: when there is base ending with 2, every 20th power will give 76 as last two digits
from 21st power same cycle repeats upto 40th power
hence till 1120, last two digits will be 76, then 1121->"02"
then 1122->"04"

Is this solution Helpfull? Yes | No (3)  

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