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What is the 28383rd term in the series 1234567891011121314............
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- 3 is the 28383rd term in the series 1234567891011121314............
There are 9 no. of single digit
there are 180 no. of double digit
there are 2700 no. of three digit
now total 2889 no. till 999
remaining no. are 25494 that is divided by 4 and the quotient is 6373 with reminder of 2 so 28381 is 6373+999=737(2)
n next no is 7(3)73
so reqd digit is 3. - 13 years agoHelpfull: Yes(41) No(10)
- 3 is the 28383rd term in the series 1234567891011121314............
There are 9 no. of single digit (1,2,3,....9)
there are 180 no. of double digit (10, 11, 12,........99)
there are 2700 no. of three digit (100, 101, 102, ..........999)
now total 2889 no. till 999
As all the no. entered from hereafter will require 4 digits
remaining no. are 25494 (28383-2889), that when divided by 4 gives the quotient 6373 with reminder of 2.
hence the series is 12345678910.....99100.......9991000.......6373-- with 2 digits still remaining to be filled. As the next no. will be 6(3)74. thus the 28382nd term is 6 and the 28383rd term will be 3. - 13 years agoHelpfull: Yes(28) No(9)
- Answer:28383..
1st term=1,2nd term=2,3rd term=3.....(n-1)th term=n-1,
so,nth term=n
therefore,28383rd term=28383 - 13 years agoHelpfull: Yes(11) No(28)
- as the given series is based on 10 so the nth term's last digit will be the digit
here the last dight will be 3 - 13 years agoHelpfull: Yes(7) No(12)
- my answer is 28383 is the 28383rd term of this series because all no.are arranged in their predefined position for eg.1 is placed at 1st, 2 is on 2nd in similar way 10 is on 10th, 11 is on 11th and so on. so 28383 will b come on 28383rd position.
- 13 years agoHelpfull: Yes(4) No(17)
- my answer is also 28383 because he said to find 28383 term because the says to find 28383 term of series but not 28383rd digit .so 1 is in 1st term 2 is second term and so on 28383 is answer
- 13 years agoHelpfull: Yes(4) No(16)
- I think the answer is 7
- 13 years agoHelpfull: Yes(4) No(26)
- Consider the series 0 1 2 3 4 5 6 7 8 9 10 11 12 13..
so up to 9 no fo digits=10
From 10 to 19 no of digits = 10*2 = 20
From 10 to 99 no of digits = 20*90= 180
So from 0 to 99 no of digits = 180+10 = 190
From 100 to 109 no of digits = 10 numbers * 3 digits = 10 * 3= 30
so from 100 to 199no of digits = 100 numbers * 3 digits = 300
From 100 to 999 no of digits = 900 numbers *3 digits = 2700
so from 0 to 999 no of digits = 2700+190 =2890
From 1000 to 1009 no of digits = 10 numbers * 4 digits = 40
From 1010 to 1019 no of digits = 10 numbers *4 digits = 40
From 1000 to 1099 no of digits = 90 numbers * 4 digits = 3600
From 1100 to 1199 no of digits = 90 numbers * 4 digits = 3600
So from 1000 to 1599 no of digits = 90 numbers * 7 *4 = 25200
So from 0 to 1599 no digits = 25200 + 2890 = 28090
no remaining digits = 28383 - 8090 = 293
From 1600 to 1609 = 40 digits
From 1600 to 1669 = 40 * 7 = 280
so from 0 to 1669 no of digits = 28090 + 280 = 28370
So from 0 to 1669 no digits = 28370
From 1670 to 1673 no of digits = 12
So from 0 to 16673 no of digits = 28382 and 28382 digit is 3
but we need 28383th term observe here from 0 to 28382 = 28383 terms
so answer= 3 - 8 years agoHelpfull: Yes(0) No(2)
- for more clearity refer >> http://it-freshers.mirchibazaar.com/accenture-sample-digit-series-puzzles
- 7 years agoHelpfull: Yes(0) No(0)
- t28383=1+(28383-1)*1
=28383 - 6 years agoHelpfull: Yes(0) No(0)
Syntel Other Question
there is a unique no of which the square and cube 2gether use all ciphers from 0 upto 9 exactly once ,which n0' is this?
) a*b*c*d*e + b*c*d*e*f + a*c*d*e*f + a*b*d*e*f + a*b*c*e*f + a*b*c*d*f = a*b*c*d*e*f and a,b,c,d,e and f are all positive nonrepeating integers then solve a,b,c,d,e, and f.