Latest TCS Aptitude Question SOLUTION: what is the remainder when (1234567890123456789)^24 divided by 6561

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what is the remainder when (1234567890123456789)^24 divided by 6561

Asked In TCS Rama Mohan Kadiri (10 years ago)

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RAKESH  (10 years ago)

ans is 0
as 123...789 is divisible by 3 and 6561=3^8

Is this solution Helpfull? Yes (26) | No (3)  

Rahul Kumar Mehta  (10 years ago)

(1+2+3+4+5+6+7+8+9+0+1+2+3+4+5+6+7+8+9) = 90

and 6561 = 3^8

so 90 is divisible by 3 thrfor rem will be 0

Is this solution Helpfull? Yes (17) | No (2)  

Dipankana Rakshit  (10 years ago)

The given numbr (123..........9)^24 =(3*x)^24 ,where x =(123....9)/3
=(3^24)*(x^24)
=(3^(8*3))*A, where A=x*24
=(3^8)*(3^8)*(3^8)*A
=(3^8)*B ,where B=(3^16)*A
=6561*B ,(since 6561=3^8)
Thus ,the given numbr(123....9)^24 is divisible by 6

Is this solution Helpfull? Yes (4) | No (2)  

Nirupam Das  (10 years ago)

It is 0 as both are divisible by 3 where 6561=3^8.

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ashish  (10 years ago)

sum of digits of denominator 6+5+6+1=18 ---> 1+8=9
sum of digits of numerator =90 =9+0=9
both nemerator nd denominator will be divisble by 9 without any remainder
i.e. 0 remainder

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Guru Pranav  (10 years ago)

It is divisible by 6561.

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