Page Numbering

$99$ pages would contribute $9\times1 + 90 \times2 = 189$ digits. Since all choices are over this and well below 3000, the last page number must be a 3 digit number. Thus, each page after 99 contributes 3 digits. Since 189 is divisible by 3, it follows that the total number of digits must be a multiple of 3. Hence the answer is 2019

$\underbrace{\overbrace{\text{Page }1\to 9}^{9 \times 1 = 9}}_{S_1=9} + \underbrace{\overbrace{\text{Page }10\to 99}^{90 \times 2 = 180}}_{S_2=9+180=189}+ \underbrace{\overbrace{\text{Page }100\to 99+n}^{n \times 3 = 3n}}_{S_3=189+3n}$

Let us defined the total number of digits $S_k$ , where $k$ is a positive integer as above. We note that the largest $\max(S_3) = 189+3(999-99) = 2889 > 2019$ . Therefore, the page number we are considering is less than 999. We note that the required $S_3 = 189+3n$ is divisible by 3 and the only option that is divisible by 3 is $\boxed{2019}$ . Then, $n=610$ and the page number is $709$ .

$99$ pages would contribute $9 + 2\times90<2000$ digits. $999$ pages would contribute $9+2\times90+3\times900>2700$ digits. Therefore, the answer must be from a book with a $3$ -digit number of pages. Thus the book has $9+2\times90+3x$ digits, which is divisible by $3$ . $2019$ is the only number divisible by $3$ , hence it is the answer.

I found a really nice way. So for the first 99 digits, the sum is: 1 + 1 + 1... + 2 + 2 +2 ... = 9 x 1 + 2 x (90 x 1) Since all this involves is 9s, the total sum is divisible by 3.

Now, for the next part, if you find the sum resulting from adding up the digits in 3 digit numbers, you'll see that it exceeds the given values. (3 x 900 = 2700) As a result, the possible value in the givens options exists within the 3 digit numbers. Now, since with every new number, you have to add 3 more, the total sum of 3 digit numbers will be divisible by 3.

In fact, the entire sum will be divisible by 3 since we deduced that for the single and double digits, the sum is divisible by 3 too. So, we just need to find out which of the 3 given options is divisible by 3.

To do so, we just use a trick by adding up all the digits and seeing if they give you a multiple of 3. With 2019, it works! (2 + 0 + 1 + 9 = 12 = 3 x 4) As a result, answer = 2019!

Well, you have 9+2x=total digits, so your total digits - 9 must be divisible by 2. 9+2x=2019,2x=2010, easily divisible by 2.

I reasoned this completely strangely. There are 9 single digit pages, 90 double digit pages, and 900 triple digit pages, etc, a "complete" book that uses all possible digit combinations is divisible by 9. So, we know that our answer, X={a,b,c} must solve the problem n*9+Xmod9=X. We also know that all pages have to have a whole number, so any remainder of dividing by nine must result in a whole number. There are only 3 answers, so it's pretty quick to note that 2019/9=224.3333, and a third of 9 gives us a whole number, 3.

Hence, 224 9+{.333 9}=2019

I feel I have explained this poorly.

there are 9 pages 1-9 (9 - 1 = 8, +1 because we're including the 1st page as well.) each with 1 number => 9 1=9 90 pages 10-99 (99-10+1=90) each with 2 numbers each => 90 2=180 9+180 gives us 189 numbers so far, less than either option so we know we have more pages. 900 pages 100-999 with 3 pages each gives us 2700 numbers, which exeeds the options so we know it's within the 3-digits pages. let's choose one of them, say 2017 and work with that. 2017-189=1828 numbers (from the 3-digits pages only) if we divide by 3 and get an integer it means this is how many numbers a certain book with so many pages contains, if it it's not suitable. 1828/3=609 remainder 1. so not 2017. to get to the next multiple of 3 we need either -1 or +2, so 2016 or 2019. 2016 isn't an option so it's 2019

I figured that the sum must be divisible by 3 since if you add the number of digits for single and double is divisible by 3 and triple digits must be divisible by three you know the solution is a multiple of 3 so 2019 is the answer

It’s last year’s number.2019

Because the number of one digit numbers is 9, two digit numbers is 90 (not including zero), and three digit numbers is 900, any number of digits given must be a multiple of nine, and so must be divisible by 3; any number divisible by three has a digital root divisible by three. Thus, of the numbers given, only 2019 (2+1+9) is divisible by 3.

709 pages would contribute 9 x1 + 90 x 2 + 610 x 3 = 2019. However, I am not sure this is correct because the number of pages should be even. A book with 1 sheet has 2 pages (1 and 2), a book with 354 sheets has 708 pages, a book with 355 sheets has 710 pages. So a right answer would be 2016.

Since they are 1 digit numbers, the sum has to be divisible by 3, and since 2019 is divisible by 3, there can be a sum of 2019.