Latest TCS Aptitude Question SOLUTION: 12341234123412....................400 digits.when divided by 909 leaves remainder ??

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12341234123412....................400 digits.when divided by 909 leaves remainder ??

Asked In TCS Vishwadeep Solpankhi (10 years ago)

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bharat kumar sharma  (10 years ago)

12341234.......400 digits
1234*10^396 + 1234*10^392 + 1234*10^388....1234*10^8 + 1234*10^4 + 1234*10^0
1234(10^396 + 10^392 + 10^388 +.........10^4 + 10^0)
now...10^0/909=1......10^4/909=1.....similarly all the powers of 10 mentioned above leaves remainder 1
1234(10^396 + 10^392......10^0)/909
1234(1+1+1+1+....100 times)and 1234/909=325 remainder
325*100=32500
32500/909=685 as remainder
thanxx to akash srivatava

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Akash Srivastava  (10 years ago)

12341234.......400 digits
1234*10^96 + 1234*10^92 + 1234*10^88....1234*10^8 + 1234*10^4 + 1234*10^0
1234(10^96 + 10^92 + 10^88 +.........10^4 + 10^0)
now...10^0/909=1......10^4/909=1.....similarly all the powers of 10 mentioned above leaves remainder 1
1234(10^96 + 10^92......10^0)/909=
1234(1+1+1+1+....25 times)and 1234/909=325 remainder
325*25=8125
8125/909=853 as remainder

Is this solution Helpfull? Yes (16) | No (17)  

Satheya Narayanan  (10 years ago)

by divisiblity prop..
1+2+3+4=10
there are 100 such sequences in 400 digits.. so 100*10=1000
1000mod909=91

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mehak jindal  (10 years ago)

ans is 3

Is this solution Helpfull? Yes (1) | No (8)  

MD  (10 years ago)

@ Bharat can u plzzzzzzz tel that (10^396........10^0) mod 909,how remainder is 1...in each ,,,,,,,,if it was 99 ,den it shud b 1,how here,plzzzzzzzzzzzz tel

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Guru Pranav  (10 years ago)

8125/909=853 as remainder

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