Latest TCS Aptitude Question SOLUTION: last two digit of 1122^1122!

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(#M40029905) TCS Numerical Ability question Keep an EYE

last two digit of 1122^1122!

Asked In TCS (10 years ago)

Solved RAKESH Read Solution (7)

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Read Solution (7) :

RAKESH  (10 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
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

Is this solution Helpfull? Yes (37) | No (4)  

Shuvadip Lahiri  (10 years ago)

1122! is an even no having a lot of zeroes at the end
so lets take 1122 ^ 100 as example
=22^100
=(2*11)^100
=2^100 * 11^100
=(2^10)^10 * 01
=76*01
=76..

Is this solution Helpfull? Yes (17) | No (2)  

rishabh raj  (10 years ago)

1122^1122! = 2^1122!*561^1122!
= (2^10)even *(01) =76

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

sindhu  (10 years ago)

2^(even num) always gives last two digits as 76

Is this solution Helpfull? Yes (2) | No (9)  

Subhabrata Roy  (10 years ago)

we know that factor of any no greater than 1 is always even.
So,22^1=22,22^2=484,22^3=10648,22^4=234256.Cyclicity of 22 is 2,4,8,6 for unit digit and 22,84,48,56 for last two digit.But we have to find 22^even no..So the last two digit is 56..

Is this solution Helpfull? Yes | No (7)  

akash  (10 years ago)

there is one formula to count successive discount that is :- P(1+r1/100)(1+r2/100). Here r1 and r2 are % of discount. after outting values in this formula we can get all values that are (a)100.40 (b)156.95 (c)150 (d)145.35. so here least price is 100.40 so answer is (A).

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Guru Pranav  (10 years ago)

76

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