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(#M40039059) TCS Numerical Ability question number system Keep an EYE

123^123! what is the late two digit

Asked In TCS Manjula (10 years ago)

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RAKESH  (10 years ago)

123^123!
(123)^(----28 zeroes)
consider only last two digit
(23)^4)^(--000000000)
(--41)^(---00000000)

unit digit = 1 ,tens place = (tens of 41)*(unit of ---000)= 4*0=0

=> last two digits = 01

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prasann  (10 years ago)

123! is obviously divisible by 4.
By power rule of 3(123) the last number must be 1 (since last digit can be 3 is 3,9,7,1)
therefore the last two digit should be X1.
Now 123! have 28 zeroes(since |123/5|+|123/25| =28) therefore it must be 4*XXUUU..0000..00.. therefore the second last digit should be X*0=0.(by Eulers form)

So the last two digits are 01

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Pranoy Bej  (10 years ago)

ANSWER==>01
as 123!= 123*122.......4*3*2*1= 4n number
so last digit will be= 1 as 3^4 =81
now 123! contains [123/5+123/25]=28 zeros
so we can write 123!=..........000000
2nd last digit will be = 2*0= 0(according to the Euler's rule 2nd last digit of power* 2nd last digit of the number= 2nd last digit)
so last two digits = 01

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lakshmi narayanan  (10 years ago)

3^1=1
3^2=9
3^3=27
3^4=81
after that all last digit is repeated by same number as 3 9 7 1
so after 4! all values are get divisible by 4! so the reminder is 0.so the last two digit is 01!!

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