Latest TCS Aptitude Question SOLUTION: the last two digits of (123)^(123!)

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(#M40030030) TCS question unsolved Keep an EYE

the last two digits of (123)^(123!)

Asked In TCS subhrajit mitra (10 years ago)

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RAKESH  (10 years ago)

123! has lots of zeroes at the end so it is divisible by 4
unit digit of (123)^(123!)=1
10th digit of (123)^(123!)=unit digit of 123 * 10th digit of 123!
=3*0=0
hence last two digit of (123)^(123!)=01

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Avik Mukherjee  (10 years ago)

ans:01..
first u know 123! have so many zeros...123/5...atlst 24...now we can easily take out 2 or 4 from this mnimum 24 zeros..
now take 123^2.....
as we know anything end with 3 we hv to tk it sqr rt..
ok 123^2 ends with 23^2..as 29..
no take 9 s power cycle and we got 29^4..as 29^2*29^2...ends with 50-29=21^2 as 41*41..ends with 50-41=9^2..as 81..
now u know a number end with 1^1245566.....ends with 1...and last two digit is very easy to count...so 81^mnoop0000000000 gives us 01 at end..


now the trick if there is a factorial in the power which is huge..u frgot about that..
then take out 4and 2 frm it..
and u can do it,,
so again frget it..
ur moto is to end the base one end with 1..which ends in 01 as a last two digit..

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shaswata tripathy  (10 years ago)

(123!)= ...............................00 its last two digits are 0s
and our question becomes 123^.......00
=__3^......00
=__9^(....00/2)
=__(-1+10)^(.....00/2)
by apllying binomial fomula we will get the 1 as first term so the last two digit will be 01
ans= 01

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saurabh agarwal  (10 years ago)

01 ans.

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kratika jain  (10 years ago)

123^123!= (3*41)^123!=(3^4)(123!/4) * (41)^123!= last 2 digit=01
3^4=81 (81)^(123!/4)=01
41^123!=01

thus 01*01=01 ans 01

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kuchu  (10 years ago)

ans is 01

Is this solution Helpfull? Yes | No (1)  

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