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123123....300 digits which div by 99 .. remainder is ??
by shyam Sundar - Saturday, 29 September 2007, 11:07 PM
 

123123....300 digits which div by 99 .. remainder is ??

 

27 18 36 33

Re: 123123....300 digits which div by 99 .. remainder is ??
by AD P - Saturday, 29 September 2007, 11:56 PM
 

123123....300 digits which div by 99 .. remainder is ??

27 18 36 33


we know that

a^n-1 is divisible by a-1

99=10^2-1

10^2n=[10^2n-1]+1
(10)^2n % 99=1
(10)^2n+1 % 99=10
(10)^2n+2 % 99=100
The given no can be written as
123[1000^299+1000^298+---------+1000^0]
(100+23)[1000^299+1000^298+---------+1000^0]
100[1000^299+1000^298+---------+1000^0]+23[1000^299+1000^298+---------+1000^0]
100[10^897+10^894+---------+10^3+10^0]+23[10^897+10^894+---------+10^3+10^0]

now we divide this term by 99
each power is a multiple of 3--so we have remainder for each term as 10
because (10)^2n+1 % 99=10
and 100 % 99=1

[10+10+10+--------+10 300 times]+23[10+10+10+----------+10 300 times]
3000+23.3000
=72000 mod 99
=27

ans must be (a) 27
Re: 123123....300 digits which div by 99 .. remainder is ??
by shyam Sundar - Sunday, 30 September 2007, 06:56 PM
  ans is 33dude
Re: 123123....300 digits which div by 99 .. remainder is ??
by fundoo bond - Monday, 1 October 2007, 03:09 PM
 

hi shyam,

    yes the ans is 33. i ll tell u my approach for such pbms...we see that the no. is divisible by 11 but not by 9.

so group the no.s such that each no. is divisible by 11. find the quotient again n then divide it by 9.

here we divide the given no. as

123123    123123    123123...............(50 such groups)

now dividing by 11,

11193      11193     11193..............(50 such groups)

so addition of all digits = 15x50 = 750

so remainder whn divided by 9 is 3.

hence total rem is 11x3 = 33.

regards, smile

fundoo

Re: 123123....300 digits which div by 99 .. remainder is ??
by lets try - Monday, 1 October 2007, 03:41 PM
  Hi,

Well given the answer options ...there is one easy way.

we can easily see that when the number is divided by 9, we get 6 as remainder....

so the final remainder when divided by 9 will give 6 as remainder..
only answer option 4th falls in above condition...
33 divided by 9 gives 6 as remainder.

Other way round number is divisible by 11, leaving 0 as remainder...so final remainder when divided by 11 will leave 0 as remainder...again option4 only....

Thanks,
Jitendra.
Re: 123123....300 digits which div by 99 .. remainder is ??
by mon haldi - Monday, 1 October 2007, 04:21 PM
 

hi jitendra

ur approach is absolutly corrt.  

but remember whenever R[a * N / a * M] is to be found

where N & M are coprime then R[a*N/a*M] = a * R[N/M}

in this case a is 33 so whatever be the ans it should be a multiple of 33

Re: 123123....300 digits which div by 99 .. remainder is ??
by shyam Sundar - Monday, 1 October 2007, 11:14 PM
 

Fundoo

How did u find the no was divisible by 11 and not 9 ?

WAsnt that arbit luck work ?? OR is there some method to the madness

Re: 123123....300 digits which div by 99 .. remainder is ??
by mon haldi - Tuesday, 2 October 2007, 01:20 PM
  hi shyam
ths is vry simple
123123 ....300digits
or (123)(123).....100times
or (1+2+3=6)......100times
so sum of all the digits are 600 this is divisible by 3 but not nine

and (123123)......50times
or (1-2+3-1+2-3 = 0)......50 times
or 0 * 50 =0 ; this shows that this is divisible by 11
isnt?
Re: 123123....300 digits which div by 99 .. remainder is ??
by aarif khan - Saturday, 6 October 2007, 04:03 AM
  hi ,
123123........300digits=123(10^297+10^294+10^291+......10^1+1)/99
rem=24(10+1+10+1+.......50 pairs)/99
      =24(550)/99
       =33
this is the coolest solution i found.
what about u?
Re: 123123....300 digits which div by 99 .. remainder is ??
by Yogendra Yadav - Saturday, 6 October 2007, 09:34 AM
 

Hi , as the test for divisibility of 9 is like this add all the INDIVIDUAL digits and then divite the sum by 9 itself and find the remainder . Thats it !

Now with the case of 99 , do the same just take digits in the group of twos and add all of them and find the remainder by dividing the sum by 99 itself .

 

here ...

 

123123123 ...... ( 300 digits )

(12 + 23 + 31) + ( 12 + 23 + 31 ) ........ So on ... ( 50 Such groups )

So .. 66 + 66 + .... ( 50 times )

=>> 66 * 50 = 3300

Now again we can do the same thing ... ( ** Plz note this step )

33 + 00 = 33

So the Remainder is 33 only !!

 

Had we were asked to reckon the remainder when the number is divided by 999 , we wud certainly have taken the groups of 3 .

... Sayonara

 

 

 

 

 

Re: 123123....300 digits which div by 99 .. remainder is ??
by aarif khan - Saturday, 6 October 2007, 08:24 PM
 

hey jitendra,
i couldn't get ur approach.
will ya plz make it more clear with another example


Re: 123123....300 digits which div by 99 .. remainder is ??
by Ranjit Paul - Tuesday, 17 September 2013, 07:13 PM
  Hi guys,

The approach for division by 99 or 999 really helped a lot. I was thinking if there is a similar kind of approach for dividing a number say 13571357..... upto 1000 digits by 101 ?
Re: 123123....300 digits which div by 99 .. remainder is ??
by TG team - Sunday, 10 November 2013, 01:39 PM
  To check the divisibility by 101 make the groups for two alternately and add them.
If the remainder is divisible by 101,the given series will be divisible by 101.
(13+13+13+13--- 500 times)= 13*500=6500
(57+57+57---- 500 times)=57*500=28500
Add them and calculate the remainder.

Re: 123123....300 digits which div by 99 .. remainder is ??
by bidyut sonowal - Sunday, 24 August 2014, 09:28 PM
  To check the divisibility by 101 make the groups for two alternately and add them.
If the remainder is divisible by 101,the given series will be divisible by 101.
(13+13+13+13--- 500 times)= 13*500=6500
(57+57+57---- 500 times)=57*500=28500
Add them and calculate the remainder


answer is 92 but it's not coming by above method naa?

plz help
Re: 123123....300 digits which div by 99 .. remainder is ??
by TG Team - Tuesday, 26 August 2014, 01:28 PM
  Hi Bidyut smile

It is mistakenly written there to 'add them'. Actually you are to subtract the two groups.

Starting from the rightmost digit of the number, first pair of number formed is to be positive and then next one negative, further next one positive and then further next one negative and so on.

So here required remainder is obtained by (57 - 13)*250 = 44*250 which is 92.

Kamal Lohia
Re: 123123....300 digits which div by 99 .. remainder is ??
by Bibs Bibs - Thursday, 4 September 2014, 12:21 PM
  Hi

If it's a big number such as 12437681243768.... upto 700 digits, then how do we find the remainder of the no divided by 101?
I get your point of subtracting but it's a little confusing here. Could you please show it's done in this case?
Re: 123123....300 digits which div by 99 .. remainder is ??
by TG Team - Thursday, 4 September 2014, 01:08 PM
  Hi Bibs smile

In earlier case we were grouping the number in 2-digit numbers from right so that each group i.e. a 2-digit number was multiple of some power of 100 (i.e. 102) and the groups were alternate +/- with first group from right most side being positive one.

Try to understand the reason on a small 4-digit number only.

Say our number is N = abcd = ab*102 + cd

Now we know that 102 when divided by 101 (i.e. 102 + 1) gives remainder -1.
So N = {ab(-1) + cd} mod 101 = {ab - cd} mod 101

If you had another group of 2-digits before ab, then that'd have been multiple of 104 which can be written as 102*102 = (-1)(-1) mod 101 = 1 mod 101. So that group would have been positive. Hence the alternate +/- pattern comes in picture here.

Now to enlarge the concept for this question, you can easily observe that in given number the pattern of digits repeat after every 7 digits. So if I consider group of 14 digits collectively, then each group would be multiple of powers of 1014..i.e.

N = 1243768
1243768*(1014)49 + 12437681243768*(1014)48 + 12437681243768*(1014)47 + ... + 12437681243768*(1014)1 + 12437681243768

Now just observe that as 102 = -1 mod 101, so 1014 = (102)7 = (-1)7 mod 101 = -1 mod 101.

Thus again we are going to get alternate +/- groups of these 14 digits, while dividing with 101, with first group from right most side being positive.

As we have exactly 50 groups of 14 digits each here and half of which are positive and half negative. Thus required remainder is simply zero as all the groups will be cancelled. smile

Kamal Lohia