## a question from Nishit Sinha

This is also a q from the same book: 12121212..........300 times/99. What is the remainder? Despite the solution is given, i am not able to understand it.

## to nishit

hi, this is about the 121212..300times problem. Since you haven't given the answer, i am not sure whether i am correct.But this is how i tried. 300 is a number divisible by 3 and is also even. so, i take the number 121212121212 [12 is also divisible by 3 and even] sum of the digits of 121212121212 = 18 which is divisible by 9 and hence 121212121212 is also divisible by 9.So, will be 121212...300 times.This follows from the divisibility rule of 9. now 121212121212/9 = 13468013468.Note 13468 is repeated twice [ no of digits/6] So we find a pattern. hence when 121212...300times divided by 9 the quotient is 134680134680134680.....013468 [ 13468 repeated 300/6= 50 times] now sum of the even digits= [3+6+0]+[3+6+0]+...= 9 *50 sum of the odd digits =[1+4+8]+[1+4+8]+... = 13*50 difference= 13*50 - 9 *50 = 4*50 =200. Thus, following the divisibility rule of 11, we get: Since, 200 is not divisible by 11 so 134680134680... (50 times) is also not divisible and hence 121212...300 times is also not divisible by 99.Hence the remainder is same as the remainder you will get when you divide 200 by 11. Hence, answer is 2 [remainder].

## thankyou sir

Thankyou sir, M.Manojkumarraju.

## hi,sorry answer shud be 2*9

hi, sorry answer shud be 2*9 = 18. remaninder should be 18. this is so as: when you divide 36/15, then you find 6 is the remainder. basically, 15= 3*5 and 36/3 = 12. now, 12/5 remainder is 2. but as we have cancelled out 3, so final remainder is 2*3=6. here 99= 11*9, but we have cancelled out 9 at the beginning. so the final remainder is 2*9 =18 and not 2. So 18 is the correct answer. in case you take up the simplier problem of 121212121212 and divide by 99 using a calculator you will find 72 as the remainder. following our method, 121212121212/9 = 13468013468 1+4+8+1+4+8= 26 3+6+0+3+6+0 = 18 26-18= 8 and hence final rmainder 9*8=72. hope i am clear now.

## hii

how cum u take the no 121212121212 not 1212121212 or 12121212 or 12121212121212.....this z da only doubt um having. howeever approach iz nice..

## actually, 121212 is the

actually, 121212 is the smallest of the numbers of the form 121212121212 . n times repeated whcih is divisible by 9. as in 121212, sum of the digits is 9. this is what came on my mind and i also noted that 6 digits is minimum and in the problem 300 is present. so i took 121212121212 so as to get the pattern of 13468013468..that's why i stated 13468 shud repeated 50 times as 300 when divided into groups of 6, the quotient is 50. got it?

## The way I solved it

Sum of 121212 = 9 and is divisible by 9 but this number is not divisible by 11. As Sum of the alternate numbers are 3 and 6 and the difference is 3. As per divisibility rule the difference needs to be 11(or 0) for the number to be divisible by 11. So we have to take eleven 12s And the sum of eleven 12s is 3x11= 33 – not divisible by 9 Next I considered twenty two 12s and the sum is 3x22 = 66 – not divisible Lastly I considered thirty three 12s and the sum is 3x33 = 99 which is divisible by both 9 and thirty three 12s is divisible by 11 So 33 x 9 = two hundred ninety seven 12s‘ll be divisible by both 9 and 11 Thus the reminder is reminder of 121/99 and that is 22

## use a calculator to find teh

use a calculator to find teh remainder. 121212/99 = 1224.36 1224*99 =121176 remainder 121212-121176= 36 and not 22. following my method 121212 divisible by 9 quotient=13468 1+4+8=13 3+6=9 13-9=4 bu cancelled out 9 hence, final remainder is 9*4=36 22 is a wrong answer. check it. don't take 121212..itz too short...take 121212121212, as else you won't understand the pattern. there is a 0 between two consecutive 13468's..right...

## for the number 121212121212

use a calculator to cross check 121212121212//99= 1224364860.72 remainder = 121212121212- 99*1224364860 =72 following my method: 121212121212/9 = 13468013468 1+4+8+1+4+8=26 3+6+0+3+6+0 =18 26-18 =8 8*9 =72[ answer u will obtain using a calculator as well]

## what i wrote earlier

what i wrote earlier

Lastly I considered thirty three 12s and the sum is 3x33 = 99 which is divisible by both 9 and thirty three 12s is divisible by 11.
Do you agree?? If yes then where is the issue..

## continuation of anita's soln

hii anita.... ur soln is correct except the last step... u tuk "digits" instead of "times".......here u got dat wrong... "Thus the reminder is reminder of 121/99 and that is " here...297 12's are absolutely divisible...rest are 3 12's so soln vil becum...(121212/9*11)...bt frm this also...um nt gettn da ansr...!!!

## nothing just confused!!

nothing just confused!!

## problem solved correct answer is 36..both were correct

following your method if u take 297's 12.. u r left with 3 12' so the number that is left out is neither 121 or 212.. rather 121212..which wqehn divuided by 99 gives 36 as the remanider were confused with total no o f digits and no of 12's 300 12's impllies 600 digits..so by mistake u took 121 i took 212..and initiallly took 50 insttead of 100 got it!!!

## i am confused may be answer is 36

i was trying to solve by your method... 12121212...297 times 2*297 -1*297 =297 divisible by 11 also sum of the digits = 3*297 ..divisible by 9 so, 121212....297 times repeated is divisible by 99 now 12121212...300 times= 121212..297tmies *1000 +212 [ not 121 as u r ending in an odd number so the last digit of 121212..297 times must be 1]right? so, it seems 121212...300 times/99 = 212/99 = 14 :)) is this ok??? but i am not finding fault in my method.. 121212121212 [ six 12's] divisible by 9 quotient =13468013468 so 13468 gets repeated twice take 121212[ 3 12's] 121212/9= 13468 take 121212121212121212 [ 9 12's] divided by 9 =134680134680134680 13468 gets repeated thrice. so i 9 12's 13468 is thrice. in 300 12's 300/3=100 times there will be so, (1+4+8)*100 = 13*100 (3+6+0) *100 =9*100 (13-9)*100 = 4*100 4*100/11 remainder is 4 4*9 remainder is 36 i am not able to find the calculation mistake..which i might have done!! can u find it..and taking smaller numbers..my method is giving correct result...

## confusing

12121212..........300 times/ is really confusing .. What does this means 12121212 (300 times) 12 (300 times) or 12 300 digits I solved with the last option but i think the 1st option is correct. Anyhow the method is not wrong

## please cross-check whetehr 36 is the corrrect answer!!

Anita's method: 121212...300 times = 121212...297 times)000000 +121212 but 121212...297 times divisible by 99 hence remainder 121212/99 which is 36 My method: 121212121212 [ six 12's] divisible by 9 quotient =13468013468 so 13468 gets repeated twice take 121212[ 3 12's] 121212/9= 13468 take 121212121212121212 [ 9 12's] divided by 9 =134680134680134680 13468 gets repeated thrice. so i 9 12's 13468 is thrice. in 300 12's 300/3=100 times there will be so, (1+4+8)*100 = 13*100 (3+6+0) *100 =9*100 (13-9)*100 = 4*100 4*100/11 remainder is 4 4*9 remainder is 36 Please corss-check

## is the answer 36? 300 12's not 300 digits hopefully

itz 300 12's ..mostprobably taking 300 12's not 300 digits... answer is coming to be 36 following both the methods cross-check

## your method was perfect

your method was perfect.... only problem both of us did, was to get confused whether itz 300 digits or 300 12's. if 300 12's then both of us are getting the same answer 36. if 300 digits, then 212/99 as i explainded..not 121 right? your method was better..mine was too conventional

## lemme give ma xplanation

guys guys guys.......its 300 digits not 300 times...hw cn u assume dat its 12 which is to b taken 300 times..it can b 1 only..it can b 121... u jst assuming urself that its 12 who has to b taken 300 times...!!!! i guess confusion is over ne.. so question becumes. 121212....300(digits)/99

## a better way to solve the 121212..problem

Question :121212...300times mod 99? The solution is simple: the given number is divisible by 9(i.e of the form 9A) and gives remainder of 4 when divided by 11.(i.e of the form 11B+4) 9A and 11B+4 are different ways of representing the same number. so equate them 9A=11B+4 .If 11B +4 is a multiple of 9 then the least value of B is 7. so the least number that is a multiple of 9 and gives a remainder of 4 when divided by 11 is 81. The next higher number is 81+99 .the next highet number id 81+2*(99) and so on...the entire family of such numbers = 99k+81. so the number in our question will also be of the form, 99k+81.So u know what the remainder is ,don't u;-)

## NOT SURE..BUT SEEMS OK

if it is 1212... (300 digits) then the answer won't be 36. 36 is ok to my feeling and senses if 12 is repeated 300 times. When 300 digits are therecomprising of 150 1's and 150 2's tnen the number is still divisible by 9 as then sum of the digits 150*3= 50*9 which is divisible by 9.now when 121216[ 6 digits] is divided by 9, then the quotient is 13468. When 121212121212[ 12 digits] is divided by 9, quotient is 13468013468.When 121212121212121212[ 18 digits] is divided by 9, quotient is 13468013468013468. Folowing this pattern it is clear that when 121212...300 digits is divided by 9, then 13468 will be 300/6=50 times. Now, sum of odd digits=(1+8+4)*50=13*50 Sum of even =(6+3+0)*50=9*50. So the difference is (13-9)*50 =4*50 =200 200/11 remainder is 2. Hence final remainder is 9*2 =18. HENCE IF 300 12'S THEN ANSWER IS 36. IF 300 DIGITS OF 1 AND 2 THEN ANSWER IS COMING 18. Not sure though but seems ok!!

## good work but ! !

and gives remainder of 4 when divided by 11
How ???? The diff. between alternate numbers is 1 so if we consider 300 digits then the sum = 150 and remainder = 7 Then as per you 9A = 11B + 7 A= 2 and B=1 and the minimum number is 18 and that’s the reminder A new method. So Guys we have 3/4 different methods for solving remainder questions. If I get some time would take all these points from this forum topic and write a book page for future reference. Wont mid if any one else does the same :-)

## i think its 300 times.

i think its 300 times. Because if it were 300 digits the question would have said so ! the answer is 36.

## 300 times of 12121212?It

300 times of 12121212? It can be but i don’t think the question maker had that in mind if that was the case then he/she would written the whole term twice or thrice before the dots.... act for me it does make any sense to write something xyxyxy. . . repeated 100 times and let the readers understand this as xyxyxy repeated 100 times because most of the time when we come across such a series we consider the repeated terms (here xy)to be present 100 times So in this case it can either be 12 repeated 300 times or in total 300 digits.

## IT IS 300 DIGITS

SEE http://www.cat4mba.com/node/1324 THIS IS THE SAME QUESTIONS AS GIVEN HERE

## see ! proves my point !

see ! proves my point !

## solution i think it gives d perfect answer

conesider 121212...300 times=1212....298 times *100 +12 add nd subtract 1212...298 times ... hence 1212..300times=1212...298times(100-1) +1212...298time +12 part before + is divisible by 99 121212...298times=121212....296times(100-1) +1212....296 times +12 hence by deducing all these we get 121212...300 times =1212 +148*12 (all d terms are removed bcoz they r multiples of 99) hence if 1212+148*12 when divisible by 99 ..........we get remainder 6........

## one more thing 1 want 2 know

one more thing 1 want 2 know is it 12 repeats 300 times r it is no. of 300 digits reply plzz nd sorry answer is not 6 its 18

## Aagain the same question

Aagain the same question ? Few think it as 300 digits and few 12 repeated 300 times, any how the way you solve it is more important

## kk if it is 12 repeatin 300 times

conesider 121212...300 times=1212....299times *100 +12 add nd subtract 1212...299 times ... hence 1212..300times=1212...299times(100-1) +1212...299time +12 part before + is divisible by 99 121212...299times=121212....298times(100-1) +1212....298 times +12 hence by deducing all these we get 121212...300 times =k(100-1)+300*12 (k is an integer) hence we have 2 find remainder wen 300*12 wen divided by 99........i.e., 36