Latest TCS Aptitude Question SOLUTION: 12341234.... 400 digits when divided by 909 leaves remainder ?

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12341234.... 400 digits when divided by 909 leaves remainder ?

Asked In TCS Shweta (10 years ago)

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Debabrata Roy Chowdhury  (10 years ago)

Ans: 685
The pattern repeats by four digits 1234!
First find the "pattern base" = 1234 (In our case!)
Then find 1234 mod 909 = 325
Now if you see 12341234 mod 909 = 325*2 mod 909 = 650
Again observe for 123412341234 mod 909 = 325*3 mod 909 = 66 and so on
So, we have 100 such "pattern base" 1234
Therefore, remainder will be given by 325*100 mod 909 = 685.

Is this solution Helpfull? Yes (34) | No (2)  

Debabrata Roy Chowdhury  (10 years ago)

Ans: 685
The pattern repeats by four digits 1234!
First find the "pattern base" = 1234 (In our case!)
Then find 1234 mod 909 = 325

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

PRATEEK JAIN  (10 years ago)

1234/909 leaves remainder 325
12341234/909 leaves remainder 650
123412341234/909 leaves remainder 0
similarly 1234123412341234/909 leaves 0 remainder
hence 1234 upto 400 digits will leave 0 remainder

Is this solution Helpfull? Yes (1) | No (3)  

pesaru karthik reddy  (10 years ago)

1234/909 leaves remainder 325
12341234/909 leaves remainder 650
123412341234 mod 909 leaves remainder 0
but this continues until 123412341234.......1234(99 times) mod 909 leaves remainder 0
therefore for 100th time we need to calculate
so 1234 mod 909=325
so answer is 325

Is this solution Helpfull? Yes | No (1)  

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