Latest TCS Aptitude Question SOLUTION: 123^123! what is the last 2 digits?

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

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

(#M40026178) TCS Numerical Ability question Keep an EYE

123^123! what is the last 2 digits?

Asked In TCS SARASVATHI (10 years ago)

Unsolved Read Solution (7)

Is this Puzzle helpful?   (2)   (1) Submit Your Solution Number System

Read Solution (7) :

Vinay Jain  (10 years ago)

Last two digits are 01

It can be find out using binomial theorem :-

123^(123)! = ((123)^2)^(123!)/2

123^2 = 15129
last two digits are 29

So we can write it as:-

((123)^2)^(123!)/2 = (-1+30)^(123!)/2

Now (123!)/2, which is a very large number and will have many trailing zeroes

we can expand the binomial expression as :-

nC0 * a^n * b^0 + nC1 * a^(n-1)*b^1 + nC1 * a^(n-1)*b^1 .... ....(no need to go further for this question)

here n = (123!)/2 , a= -1 b = 30

Putting values:-

(123!)/2 C0 * (-1)^(123!)/2 30^0 + (123!)/2 C1 * (-1)^((123!)/2 -1) * 30^1
| | |
| | |
| | |
| | |
v v v

last digit: 1 + last digit: 000000 + last digit: 000000

So final addition will be like .....000001

Last two digit 01 Ans

jisko samajh nhi aya wo call krlo muje 09981241341, smjha dunga :P

Is this solution Helpfull? Yes (9) | No (3)  

vanika gakwad  (10 years ago)

ans 67
123^123 convert it such that its last digit is 1
123^120*123^3=(123^4)^30*123^3=(41)^30(last 2 digits)*123^3
41^30 last 2 digit= 4*0=0 & 1 i.e. 01 *123^3=67 ans

Is this solution Helpfull? Yes (6) | No (5)  

Diksha   (10 years ago)

last two digits: 01.

explanation : 1. check the cyclicity of 123 that is 4
123^123! = (123^4)^xyz000000....
therefore 01

Is this solution Helpfull? Yes (5) | No (8)  

anu maurya  (10 years ago)

(123^4)^xy000.. nw apply cyclicity rule answer will be 01

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

ayush gupta  (10 years ago)

ans is 67...

Is this solution Helpfull? Yes | No (5)  

Prateek Pandey  (10 years ago)

since last 2 digits are needed so:
123^123! mod 100
Carmichael Number for 100 is 20
Since 20 divides 123!, so answer is to be unity
Hence 01

Is this solution Helpfull? Yes | No   

Guru Pranav  (10 years ago)

123^..........0000...28 zeroe's

==01 as last two digits !!!

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.