Latest TCS Aptitude Question SOLUTION: Find the last two digits of 2025^2052+1392^1329

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Find the last two digits of 2025^2052+1392^1329

Asked In TCS dipali (10 years ago)

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ABDUL ODUD  (10 years ago)

Ans is 37 .
1392^1329-------> 2^1329 or (2^10)^132 *2^9 or 76*12=12
and 1st part --->last 2 digit is 25
so ans 25+12=37
thanx :)

Is this solution Helpfull? Yes (15) | No (10)  

sameer singh  (10 years ago)

first part is 25 always
second part is 72
thus total is 97

Is this solution Helpfull? Yes (10) | No (1)  

Debolina Chakraborty  (10 years ago)

1st part is always 25.
2nd part implies
23^1329*2^2658 implies
23*(23^4)^332*2^8*(2^10)^265 implies
23*41^332*256* 24^265 implies
23*81*256*24 implies 72
nd 25 + 72=97..

Is this solution Helpfull? Yes (7) | No (3)  

mehak jindal  (10 years ago)

25+32=57 ans

Is this solution Helpfull? Yes (4) | No (7)  

Ashish Prajapati  (10 years ago)

2025^2025+1392^1392
25^2025+92^1392
25+(2*2*23)^1392
25+96
ans=21

Is this solution Helpfull? Yes (2) | No (7)  

pritam banerjee  (10 years ago)

give a nice logic ...

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

Shantanu Mukhopadhyay  (10 years ago)

2025^2052 + 1392^1329
=> 25^2052 + 92^1329
=> 25 + (26 x 12)
=>25 + 88
=>13

ans is 13.

Is this solution Helpfull? Yes (1) | No (7)  

Totan Shit  (10 years ago)

2025^any digit =last digit will be 5
1392^1329=(1392^1325)*(1392^4)=last digit will be 2
so 2025^2052+1392^1329 =lat digit will be 7

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

Rudra  (10 years ago)

what is the actual logic when we have to find the any power of any no. plzzz anyone explain in detail....

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Karthika  (10 years ago)

Can anybody clearly explain how 72 is obtained as the last 2 digits for 1392^1329 ??

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Karthika  (10 years ago)

@Rudra:
See if d last digit of 45^2112 is asked,the answer is 5
Similarly if the last digit of 27^5001 is asked,yu need to do the following steps
we know the last digit of 7*7*7*7 is "1"..i.e the last digit of 7^4 is 1. similarly last digit of 7^8=1..and so on 7^(any multiple of 4)=1;
So we can write 27^5000 as (27^(4*1250)).here the last digit is of the result is 1.
but we need 27^5001, so (27^(4*1250))*27. hence 1*7(last digit of 27)=7

So now yu shuld be able to find the last digit of 67^5002.similarly

I find it easy to find the last digit of power questions. But am not sure about the last 2 digits.!! Anybody please explain it to us. How do we find the LAST TWO DIGITS of a greater power arithmetic problem?

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Shantanu Mukhopadhyay  (10 years ago)

2025^2052 + 1392^1329
=> 25^2052 + 92^1329
=> 45 + (26 x 12)
=>45 + 88
=>33

ans is 33.

Is this solution Helpfull? Yes | No (23)  

Danish Shaikh   (10 years ago)

5^anything gives 5 , similarly 2^anything gives 2
05+02=02

Is this solution Helpfull? Yes | No (17)  

ayushi garg  (10 years ago)

plz explain the logic of this question..

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deepak jain  (10 years ago)

Ans. is 97.
as first part is 25 surely and second part is giving 72..
so its 97.. its coming mathematically.

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akhil reddy  (10 years ago)

plz send the correct ans explanation to this question

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