Latest TCS Aptitude Question SOLUTION: what is the remainder when 123412341234.....upto 400 digits is divided by 909?

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(#M40026389) TCS Numerical Ability question Keep an EYE

what is the remainder when 123412341234.....upto 400 digits is divided by 909?

Asked In TCS pre (10 years ago)

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pranay sharma  (10 years ago)

1234%909=325
12341234%909=325*2
1234........400 digits=325*100
ans=(325*100)%909 = 685

Is this solution Helpfull? Yes (58) | No (2)  

Prajul laad  (10 years ago)

685 is d correct ans

Is this solution Helpfull? Yes (12) | No (2)  

Ramil Gogna  (10 years ago)

@puneet: please explain

Is this solution Helpfull? Yes (3) | No   

Rituparna Bhattacharyya  (10 years ago)

ans :-> 325

123412341234 is completely divisible by 909
400%909 = 4
1234 % 909 = 325

Is this solution Helpfull? Yes (3) | No (8)  

puneet panwar  (10 years ago)

1
it is done using g.p. series .

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

Gunjan Khatter  (10 years ago)

D ans z 315

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

puneet panwar  (10 years ago)

@ramil:
first step 1234(10^396+10^392+10^388.......+1)
now a g.p. is formed inside with a=1 and r=10^4
solve for the value of no. of terms in g.p n=100
find sum of g.p , it will come equal to 999999.......(400times9)/9999
in next step divide by 101 and you will get {1234*(1010101010.......1)}/9*101
now 1010101...1 divisible by 101 so you will get remainder by dividing 1234 by 9 and that is equal to 1.

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Guru Pranav  (10 years ago)

1234%909=325
12341234%909=650=325*2
.
.
.
1234 ..... 1000 times = (325*100)%909=685 Ans.

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UMAMAHESWARI.P  (10 years ago)

1234(4 digit),400/4 gives reminder 0 and q-100
so the reminder 0

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Heena Jain  (10 years ago)

PRE plz provide us the options..

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sangram badi  (10 years ago)

my be 200 ???

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ARISH KHAN  (8 years ago)

1234%909=325
12341234%909=325*2
1234........400 digits=325*100
ans=(325*100)%909 = 685

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