Latest TCS Aptitude Question SOLUTION: what is the last two digits of 123^123!

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what is the last two digits of 123^123!

Asked In TCS gaurav (11 years ago)

Solved Asit Shukla Read Solution (11)

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Read Solution (11) :

Asit Shukla  (11 years ago)

123! will have more than 2 zero at the end, so it is a multiple of 4.

Now (123^4)^(...00)
=(228886641)^(...00)
Now powers of 41 repeat the last two digits.
ie 41^1=41 41^6=..41
41^2=..81 41^7=..81
41^3=..21 41^8=..21
41^4=..61 41^9=..61
41^5=..01 41^10=..01

As 123! will still be having some zeroes left at the end so it value is a multiple of 10 and so the last 2 digits are 01.

Is this solution Helpfull? Yes (45) | No (7)  

Swapnasish Dutta  (11 years ago)

bt 123! is on the power,so how it will be 00 ?? @Piyush

Is this solution Helpfull? Yes (18) | No (3)  

guruteja  (11 years ago)

it is in the form
(123)^(x0000..(28 zeros))
=>((123)^4)^(y000..)
=> (z41)^(y00..)
=>units place=1,tens place=4*0=0
=>last two digits=01.

last two digits will be 01

Is this solution Helpfull? Yes (11) | No (7)  

Piyush Kumar Chauhan  (11 years ago)

The last two digits should be 00 because on calculating the factorial of 123, we have 1 * 2 * 3 * ....... * 123, so we get 2 * 5 and 10 also in between so their product is 100 whatever else multiplied the last two digits will be 00 only.

Is this solution Helpfull? Yes (9) | No (18)  

Amandeep Kaur  (11 years ago)

mr.asit shukla....can u elaborate the soln...so that it becomes more clear..

Is this solution Helpfull? Yes (5) | No (1)  

subrata kumar  (11 years ago)

last 2 digits of any no. A^(40k+r) = A^r...
in this case 123^123= 123^((40*3)+3)= 123^3 = 67

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

hardik  (11 years ago)

mr. asit shukla can u please elaborate more esailyyy

Is this solution Helpfull? Yes (4) | No (3)  

seetha  (11 years ago)

the last digit will be 9 bt dono about last but one digit

Is this solution Helpfull? Yes (3) | No (10)  

Sneha Yadav  (11 years ago)

last two digits will be 01

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

Guru Pranav  (11 years ago)

01 will be the last two digits !!!

Is this solution Helpfull? Yes (2) | No   

Utsav Chakraborty  (10 years ago)

as 123! will be divided by 4. So (123^4)^(.......00) because 123! will have atleast two zeroes in the end digits.
Now 123^4=228886641.Here 41 is the last two digits as
228886641^(.............00)
So by checking the power cycle of 41,
41^1=41
41^2=81(note:we are checking only last two digits so only last two digts i am writing)
41^3=21
41^4=61
41^5=01
and 41^6=41 from this stage the last two digits starts repeating itself.
So the last two digits are 01.

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

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