Latest TCS Aptitude Question SOLUTION: what is the last two digits of 1122^1122!

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Number System

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40039088) TCS Numerical Ability question number system Keep an EYE

what is the last two digits of 1122^1122!

Asked In TCS vinitha (9 years ago)

Unsolved Read Solution (12)

Is this Puzzle helpful?   (17)   (4) Submit Your Solution Number System

Read Solution (12) :

Atchut KUmar  (9 years ago)

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
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

Is this solution Helpfull? Yes (15) | No (4)  

Swayanti Bhaumik  (9 years ago)

76 it is
2^10^even will end with 76
Now 1122! Will be an even number only...taking 10 common will result in even number...
Thus 2^10^even= 76

Is this solution Helpfull? Yes (8) | No (1)  

muthukumaran  (9 years ago)

cyclicity of 2 is 4
factorial value will be divisible by 4 and remainder is 0.
number at the place of 0 in the cyclicity is 6

Is this solution Helpfull? Yes (5) | No (3)  

saketh  (9 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 (5) | No   

Roshni  (9 years ago)

1122*(1000+122)
=1122000+1122*122
1258884

Is this solution Helpfull? Yes (4) | No (11)  

gautam gaurav  (9 years ago)

ans will be 76

Is this solution Helpfull? Yes (3) | No (1)  

prashansa  (9 years ago)

76 coz 2^10^even and 1122! will be even.

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

bavireddygowtham  (9 years ago)

84 last two digits means 22^1122
22 we can write as (2*11)^1122 2^1122*11^1122= 4*21=84

Is this solution Helpfull? Yes (1) | No (2)  

nupur israni  (9 years ago)

any even no raised to the power of 20k ends with 76 as its last two digit and any odd no raised to the power of 20k ends with 01 as its last two digit.. since 1122! will be a multiple of 20 (1*2*......*19*20*.......*1122) and 1122 is an even no therefore the ans will be 76.

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

Atchut KUmar  (9 years ago)

May be 6 I guess..

Is this solution Helpfull? Yes | No (5)  

pooja verma  (9 years ago)

1122!is divibile by 4n so
2^4 = 6 last digit is 6

Is this solution Helpfull? Yes | No (2)  

Damini Chaurasia  (9 years ago)

16
1122=11*102
11^1122! * 102^1122!
1122! contains some zeroes at the end..so 11^1122!=01 (last 2 digits)
1122! is divisible by 4 so we are now left with (102)^4=16 (last 2 dgts)

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.