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Logical Reasoning
Number Series
What is the remainder when 121212..(300 digits) is divided by 99?
a. 18
b. 36
c. 72
d. 0
Read Solution (Total 9)
-
- 1) 99 can be expressed as 9*11
2) divisibility rule for 9: sum of digits must be divisible by 9
so,pair 2 digits. 150 total pairs and sum of digits of each pair is 1+2=3 toal sum is 3*(150)=450.
450 is completely divisible by 9. rem=0
3) divisibility rule for 11: sum of evn places - sum of odd places must be 0 0r 11 multiple.
consider 1st 2 digits 2-1=1
similarly 150 pairs*1= 150
150/11= rem is 7
4) we can express it as 9A=11B+7
give A,B=0,we will get 9 7
give A,B=1,we will get 18 18
we should give same values for A and B untill we get equal values and that will be answer. so answer to this is 18 - 11 years agoHelpfull: Yes(17) No(7)
- 18 iska bhi nai bataunga...... hehe
- 11 years agoHelpfull: Yes(6) No(40)
- ans:18
[1212..(300 digit)]%99=[10..(299 zeroes)+20..(298 zeroes)+...+200+10+2]%99
=[.......+10+2+10+2](as 2%99=2,10%99=10,200%99=2,1000%99=10,20000%99=2 so there is a pattern : remainder=2 when digit is 2... and remainder=10 when digit is 1...)
now 10 is 150 times and 2 is also 150 times= 10*150+2*150=1800.
now again 1800%99=18 ans - 11 years agoHelpfull: Yes(6) No(1)
- 12%99=12
1212%99=24
121212%99=36
1212..... 300 times % 99= {12 * (300/2)} % 99= 18(ans) - 11 years agoHelpfull: Yes(6) No(0)
- Here in this question 12 comes 150 times ...... If we add all the digits , we get 4500 which is divisible by 9 . Now we divide 99 by 9 , we get 11 since 99 = 9*11. So remainder has to be below 11 . The correct option has to be 0. S.I.T.Tumkur Royal Mech
- 11 years agoHelpfull: Yes(4) No(8)
- (121212121.....12)/(3*3*11)
the above number is divided by 3 gives remainder=0
remainder is 0 - 11 years agoHelpfull: Yes(2) No(2)
- 36
logic only
- 11 years agoHelpfull: Yes(1) No(5)
- 7 as it is completely divide by 9 and wen / by 11 it leaves remaainder 7.
- 11 years agoHelpfull: Yes(0) No(1)
- 0 zero is the answer, becoz its completely divisble !!!
- 11 years agoHelpfull: Yes(0) No(1)
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