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number system problems
by dr naina singh - Saturday, 12 July 2008, 04:45 PM
  1.if you for a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of nine, what can be the maximum number of elements in the subset
a)1668   b)1332  c)1333  d)1334


2.find the 28383rd  term of series 12345678910111213141516...........
a)3    b)4    c)9    d) 7


3. The product of the digits of a two digit number is twice as large as the sum of its digits . if we subtract 27 from the required number, we get number consisting of the same digits written in the reverse order . find the no?



plz post the approach.
Re: number system problems
by TG Team - Saturday, 12 July 2008, 05:58 PM
  3. The product of the digits of a two digit number is twice as large as the sum of its digits . if we subtract 27 from the required number, we get number consisting of the same digits written in the reverse order . find the no?

10a + b - 10b - a = 9(a - b) = 27
=> a - b = 3
or a = b + 3.
Also, ab = 2(a + b)
=> b(b + 3) = 2(2b + 3)
=> b2 + 3b = 4b + 6
=> b2 - b - 6 = 0
=> (b - 3)(b + 2) = 0
=> b = 3, because b > 0
=> a = 6
and the number = 63.smile
Re: number system problems
by TG Team - Saturday, 12 July 2008, 06:32 PM
  2.find the 28383rd  term of series 12345678910111213141516...........
a)3    b)4    c)9    d) 7

one digit numbers from 1 to 9 = 9
two digit                     10 - 99 = 90
3 digit                       100 - 999 = 900
4 digit                      1000 - 9999 = 9000

Now 28383 = 9000 + 900 + 90 + 9 + 3676*5 + 4.
Hence 28383rd term of the series is 4th digit of 3677th 5 digit number starting from 10000. i.e. 13677.
Hence answer is 7 smile
Re: number system problems
by dr naina singh - Saturday, 12 July 2008, 08:24 PM
  kamal can u plz explain how can u deduce that it is  4th digit of 3677th 5 digit number starting from 10000. sorry for bothering u 
Re: number system problems
by balakrishna t - Saturday, 12 July 2008, 10:10 PM
 

1. 1334

You can take all the even nos - the nos which are even multiples of 9 (i.e 18, 36 ....)

Since we have 5 no s in 90 in 3000 we have 166 such nos

So 1500- 166 =1334

Am i correct ??? confirm

3. Ans : 63

 

Regards

Balakrishna

 

Re: number system problems
by dr naina singh - Sunday, 13 July 2008, 12:26 PM
  sorry balakrishna t for ans no 1 u have to try again . ans is not 1334
Re: number system problems
by sanki shaitan - Sunday, 13 July 2008, 01:50 PM
  ans 1)  (c)1333
Re: number system problems
by dr naina singh - Sunday, 13 July 2008, 11:55 PM
  yes sanki shaitan u got it . but can u post the approach u follow 2 solve it.
Re: number system problems
by rashi agarwal - Monday, 14 July 2008, 12:22 AM
  3.
let the no. be 10x+y

so,10x+y-27=10y+x
or x-y = 3

here use some hit and trial,
  x y
  2 7 +
= y x.
therefore x<=7. the possible pairs are (7,4),(6,3),(5,2) and )(4,1).because x-y=3.
after checking the required condition answer is 63.

Re: number system problems
by Top CAT - Monday, 14 July 2008, 07:10 AM
   Hi Kamal

2.find the 28383rd  digit of series 12345678910111213141516...........
a)3    b)4    c)9    d) 7
                                                                  total no. of digits
one digit numbers from 1 to 9 = 9                                       9
2 digit                         10 - 99 = 90                          2*90   = 180 
3 digit                       100 - 999 = 900                       3*900 = 2700
4 digit                      1000 - 9999 = 9000                   4*9000 = 36000

Now 28383 = 9 + 180 + 2700 + 25494
Now 25494 =  6373 * 4 + 2
that means 6373 th term after 9999 which is 16372 and next term will be 16373 so 28383rd  digit will be 6

 Hi naina tell me if i am right
Re: number system problems
by Top CAT - Monday, 14 July 2008, 07:27 AM
  1.if you for a subset of integers chosen from between 1 to 3000, such that no two integers add up to a multiple of nine, what can be the maximum number of elements in the subset
a)1668   b)1332  c)1333  d)1334




we can take any no of series 9x+1,9x+2,9x+3,9x+4

if we add any two nos we won't get a multiple of nine

{1,2,3,4, 10,11,12,13 19,20,21,22 .............2989,2890,2891,2892, 2998, 2999 ,3000}

for every 9 nos there are 4
total nos mod3000/9 * 4 = 1332 + 3 other nos (2998,2999,3000)
plus now we can add any multiple of 9 to this subset

therefore total nos 1336

Re: number system problems
by CATendra 2008 - Monday, 14 July 2008, 10:19 AM
  @TOP CAT
that means 6373 th term after 9999 which is 16372 and next term will be 16373 so 28383
rd  digit will be 6

till 9999 we already have 36000 digits.Don't we?
I think it will be 6373th term after 999 which makes it 6373.Hence,28383rd digit is 3
Please correct me if I am wrong.
Re: number system problems
by rashi agarwal - Monday, 14 July 2008, 04:32 PM
  hello topcat,
First of all thanx for the wonderful solution.I jst want to say that if we exclude 1 and 3000 then the solution will be 1334. so please explain  if 1 and 3000 will be included or not?
Re: number system problems
by Amit wadhwa - Monday, 14 July 2008, 05:01 PM
 

hi CATendra..

digit will be 6 only bcoz 28383 = 6373*4+2( this 2 signify the second digit of the number from right) and 6373 th term after 9999 which is 16372(9999+6373) and next term will be 16373 so 28383rd  digit will be 6.. ie second digit...

hope this will clear your doubt..

Re: number system problems
by suresh kumar - Monday, 14 July 2008, 08:32 PM
 

hello naina singh

please tell me are the sol provided  below correct.......

prob.1

as we know considering 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,......

u will find that only four of the digits b/w 1 to 9 or 10-18 and so on..... are consistant not to make multiple of 9 when taken in pairs like1,2,3,4,10,11,12,13,19,20,21,22,........

so no of  sets will be (3000/9)=333+3/9(sets of nine digits...)

no of digs =333*4=1332+3(last three digits of the seqience 2998,2999,3000)=1335

now add one dig a 9 multiple

therefore1336

 

prob.2

12345678910111213141516..

its simply counting of dig

digits in one dig no.=9

digits in two dig no.=90*2

digits in three dig no.=900*3

 digits in four dig no.=9000*4

9+180+2700+36000=38889

it means above digit is in four digit no(any)

calculate as

28383-(9+180+2700)=25494

four dig no will be..25494/4=6373+2/4

therefore this series will start after 999 and 7372 will be this no and as 2 is remainder 3 will be the required digit

prob.3

probem 3 is guessing

the diff in no is 27 this is possble in case of

14,25,36,47,58...thereno here is 36..

 

Re: number system problems
by dr naina singh - Monday, 14 July 2008, 09:59 PM
  Ans r
1)1333
2)9 but i am getting 7
3)63
 TG pleeeeeeeeeeeeeeeeeeeeeeezzzzzzzzzz through some light on ques no 1 and 2 .
Re: number system problems
by Top CAT - Tuesday, 15 July 2008, 05:56 AM
  Hi Rashi
If you for a subset of integers chosen from between 1 to 3000 (not including)
hence we should not include them
 so ans is 1334
Re: number system problems
by nitin . - Thursday, 17 July 2008, 11:55 AM
  hi Topcat Plz let me know why you have added these 3 nos again.

total nos mod3000/9 * 4 = 1332 + 3 other nos (2998,2999,3000)
Re: number system problems
by Top CAT - Thursday, 17 July 2008, 08:47 PM
  hi nitin
there are 1332 nos from 1 to 2892 in the form 9x+1,9x+2,9x+3,9x+4
after them there are still other three nos ie 2998,2999,3000 in the form 9x+1,9x+2,9x+3
Re: number system problems
by sanki shaitan - Saturday, 19 July 2008, 08:33 AM
  hello naina;
the answer to your first question can be approached as follows-

from 1 to 3000 all the numbers can be written in the form 9k,(9k+1),(9k+2),.....,(9k+9)

now divide 3000  by 9 therefore we get 333 as quotient.
further we have to exclude 1 and 3000 as numbers are asked between 1and 3000.
therefore numbers of the form-
9k=333+the number 2998-the 1 which we have counted and has to be excluded
9k+1=333+the number 2999
9k+2,9k+3,9k+4,9k+5,9k+6,9k+7,9k+8,9k+9=333

now we have to choose a subset of integers such that no two integers add up to a multiple of nine. therefore we cannot consider 9k+9. further out of (9k+1 and 9k+8 ).(9k+2 and 9k+7 ),(9k+3 and 9k+6 ) and (9k+4 and 9k+5 )

now we have to to find the max number of elements in the set therefore we have to calculate
max(elements of the form 9k+1 or 9k+8 )+max(elements of the form 9k+2 or 9k+7 )+max (elements of the form 9k+3 or 9k+6 ) + (elements of the form 9k+4 or 9k+5 )
=333+334+333+333
=1333

hope the solution is clear....
evil

Re: number system problems
by niraj srivastava - Sunday, 20 July 2008, 10:35 AM
 

answer

1)1334

2)3

3)63

Re: number system problems
by vishal bawa - Friday, 25 July 2008, 08:28 PM
  dear amit...what i feel is tha your concept is right...but what i feel is that you have done i silly mistake...it should be 6373 term after 999 not after 9999....so this makes it 7372 and thus 28383 term shal be 3 of 7373 digit....
plz correct me if i am wrong
Re: number system problems
by Top CAT - Friday, 25 July 2008, 08:44 PM
  Hi vishal
thanks for pointing out that mistake

6373rd term after 999 will be 6372 and next term will be 6373 so 28383rd  digit will be 3

jinesh
Re: number system problems
by Varun Vashishth - Monday, 4 August 2008, 11:07 AM
 

Dude...me gettin 1335 (same as yours and by applying same concept..)

did u get any cnfirmation..?

Re: number system problems
by Karan Singh - Sunday, 6 September 2015, 09:23 AM
  Its wrong.