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Number with two or more digits in which the digits reading left to right occur in strictly increasing order are called as "sorted numbers". For example 125, 14 and 239 are sorted numbers while 22, 74 and 198 are not sorted. Suppose that a complete list of sorted numbers is prepared and the numbers themselves are written in increasing order, what will the 100th sorted number on the list be?

The options are 389, 356, 345, and 258.

Can please anyone solve this and give me an explained answer and concept used?

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    $\begingroup$ Please rewrite the question in proper english so that we can actually understand what it is about. Also provide context about what you have tried, what you know about the problem and where you get stuck. $\endgroup$
    – Ove Ahlman
    Commented Mar 29, 2016 at 13:30
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    $\begingroup$ Maybe $25$ is sorted number? $\endgroup$
    – openspace
    Commented Mar 29, 2016 at 13:31
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    $\begingroup$ 25 seems a sorted number, according to the definition geven before. Did you mean '22 is not sorted'...? $\endgroup$
    – CiaPan
    Commented Mar 29, 2016 at 13:33
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    $\begingroup$ I would start by working out how many one-digit numbers are sorted and how many two-digit numbers are sorted. You will then have a good "road map" to discovering how far into the three-digit numbers one needs to go to find the hundredth sorted number. $\endgroup$
    – hardmath
    Commented Mar 29, 2016 at 13:39
  • $\begingroup$ yes @CiaPan 22 is not sorted $\endgroup$
    – prog_SAHIL
    Commented Mar 29, 2016 at 13:40

4 Answers 4

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12-19 are sorted; 20, 21, 22 are not. So we have 8 sorted numbers, 3 unsorted (where 1 is the first digit)

23-29 are sorted; 30, 31, 32, 33, 34 are not. So we have 7 sorted numbers, 4 unsorted (where 0 is the first digit)...

And we continue this pattern of decreasing sorted numbers until 89, since the 90's have no sorted number. So 8+7+6+...+1 = 36 sorted numbers.

There is not another sorted number until 123, 124, 125, 126, 127, 128, 129 (7 sorted)

And the 130's would have 6 sorted, etc... Another 7+6+5....+1 = 28 sorted numbers.

And in the 200's, there will be 7+6+5...+1 = 21 sorted

300's; 6+5+4+...+1 = 15 sorted.

This brings us to the 100th sorted number, since 36 + 28 + 21 + 15 = 100.

So the last sorted number in the 300's, is equal to 389.

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  • $\begingroup$ Thanks for the method. But how can we exactly find the 100th sort number??? $\endgroup$
    – prog_SAHIL
    Commented Mar 29, 2016 at 13:52
  • $\begingroup$ I have edited it. $\endgroup$
    – Inazuma
    Commented Mar 29, 2016 at 13:56
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    $\begingroup$ upvote ^^. Thank you so much. $\endgroup$
    – prog_SAHIL
    Commented Mar 29, 2016 at 13:58
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In any permutation of $n$ distinct digits, there is just one sequence of strictly increasing digits, thus the number of combinations of $n$ distinct digits = # of sorted numbers

There are $\binom92 = 36$ sorted two-digit #s

There are $\binom82 = 28$ sorted three-digit #s starting with $1\to 64$ running total

There are $\binom72 = 21$ sorted three-digit #s starting with $2\to 85$ running total

There are $\binom62 = 15$ sorted three-digit #s starting with $3\to 100$ running total

Thus $389$ should be the hundreth sorted #

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Not sure if there is an easier way to do this, but I broke this up into sets. For the two digit numbers:

{12, 13, 14, ..., 19} : there are 8 sorted numbers with a leading 1 {23, 24, 25, ..., 29} : there are 7 sorted numbers with a leading 2

Carrying out this logic, there are 8 + 7 + ... + 1 = 36 two digit sorted numbers.

Next, look at three digit sorted numbers. There are 7 + 6 + ... + 1 = 28 three digit sorted numbers that start with 1, 6 + 5 + ... + 1 = 21 three digit sorted numbers that start with a 2, and 5 + 4 + 3 + 2 + 1 = 15 three digit sorted numbers that start with a 3.

So, I got the same answer as Archis - 389

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  • $\begingroup$ true blue anil's answer is a better approach $\endgroup$
    – ashleydc
    Commented Mar 29, 2016 at 14:13
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Note the peculiarity there are $36$ such numbers ie (8 in $12-19$,7 to $23-29$,...1 in $80-89(89)$) in $1-100$ then $28=36-8$ such numbers in $100-200$ ie ($123-129$ have 7... $180-189$have 1) ... And so on ...so we want 100th number so 100=36(total numbers in($1-100$)+28(total numbers in $100-200$)+21(total numbers in ($200-300$)+15(total numbers in $300-400$) so its the last number of the range $300-400$ which is $389$

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  • $\begingroup$ Could you clarify your answer? I am not understanding the method you are using, nor how you reach the answer. $\endgroup$
    – shardulc
    Commented Mar 29, 2016 at 13:46
  • $\begingroup$ i am not getting how did you found there are 36 numbers in 1-100 and 28 in 100-200 $\endgroup$
    – prog_SAHIL
    Commented Mar 29, 2016 at 13:53
  • $\begingroup$ Edited any problem now $\endgroup$ Commented Mar 29, 2016 at 13:53
  • $\begingroup$ marked it correct. $\endgroup$
    – prog_SAHIL
    Commented Mar 29, 2016 at 13:54

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