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If it is given that a particular digit is used 'n' times in the page numbers of a book, can we definitively say how many pages the book have ?And if it is possible to determine the number of pages in which that particular digit appears in the page numbers

For example in one of the questions, it was asked to find the possible number of pages in the book ( out of 1024, 1124, 1134, 1224, 1234) if the digit '1' was used 689 times in the page numbers of the book.After eliminating all the options we got the solution at 1234.

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    If the digit $1$ is used once, can you say how many pages are in the book? What about if used twice?2017-02-25
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    @Mike thing is these are consecutive pages. So, the digit 1 is obviously used multiple times when pages cross double digits. And, to your question, I think the answer is yes2017-02-25
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    Assuming the pages start at 1, and all pages are numbered in order, then you can predict a lot. You really need to play with this on your own. But how many times will it appear in the thousandth place?2017-02-25
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    So if the 1 is used exactly once, did the book cross double digits? If 1 is used exactly 2 times did it?2017-02-25
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    It might be easy to ask: if the book has nine pages, how many times is each digit used. If the book has 99 pages, how many times? If the book has 999? So how can a book have a number in between?2017-02-25

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If a book has 9 pages each digit is used exactly once. If a book has 99 pages, each digit is used 10 tines in the ones place and ten times in the tens place or 20 times. If the book has 999 pages each digit is used in the ones place 100 times, the tens place 100 times, and the hundreds place 100 times or 300 times.

So if the book has 9999....9 = $10^k-1$ pages, each digit is $10^{k-1} $ times in each of the $k $ positions so each digit is used $k*10^{k-1} $.

So if, say, the number $5$ is used $7349$ times then:

$4000 < 7349 < 50000$ so there are between $9999$ and $99999$ pages. The digit $5$ was used $4000$ up to page $9999$ so the remaining $3349$ were used in pages $10,000$ through $99,999$.

Between $10,000$ through $19,999$ each digit, except 1, appears another $4000$ times so there fewer than $19,999$ pages.

Between $10,000$ and $10,999$ each digit except 1 and 0 appear 300 times. Between $11,000$ and $11,999$ another 300 times. So between $10,000$ and $14,999$ the digit $5$ occurs $5*300=1500$ times.

We still have $1849$ occurrences to account for.

From $15,000$ the $15,999$ the digit $5$ has occured $300$ times in the ones, tens, and hundreds place, just like before, in in the thousands place it has occured $1000$ times. This accounts for $1,300$ more times. We have $549$ more occurences to account for.

Pages, $16,000$ to $17,000$, again account for $300$ and $17,000$ to $18,000$ will account for $300$ more. So we know the book has between, $17,000$ and $17,999$ pages.

We have $249$ more occurences to account for. Between $17,000$ and $17,100$ each digit (except 1 and 7) occur $20$ times so between $17,000$ and $17,499$ the digit $5$ occurs $5*20=100$ times. We have 149 occurences to account for. Between $17,500$ and $17,599$ the $5$ occurs 20 times in the one and tens place, but $100$ times in the hundreds. So that accounts for $120$ more and we have $29$ more to go.

$17,600$ to $17,700$ account for $20$ more leaving just $9$. Between $17,700$ and $17,749$ there are $5$ occurences in the ones position. $4$ more to go.

$17,750-17,753$ account fo the last $4$ occurences.

The book has exactly $17,753$ pages.

Note, not all numbers have answers. $7350$ would put as at $17,754$ pages but $17,755$ pages makes the number $7352$. The number $7351$ is impossible.

Also not all numbers have an exact number of pages. If there were say, exactly $3$ of the digit $5$, then we could have had anywhere between $25 - 34$ pages.

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    I would like to point out a small error in your analysis that if the book has 99 pages then all the digits appear 20 times except 0, which appears 9 times. Anyways, I think we cannot have a particular solution to all such questions unless some other conditions like the no of repeatation of some other digits are given2017-02-25
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    Yeah, I did notice that about half way through. I think we can have a solution although the notation would be a nightmare so that a description would be easier.2017-02-25