1000 page book, without pages with odd digits

Question

In a book of a thousand pages, we rip off those which have numbers with at least one odd digit, i.e. we rip off $768$ and $100$ but we don't rip off $248$.

a. Which page is in the position $100$ after the pages are ripped off?

b. Which position holds the page with the number $888$?

I started writing a generic number in decimal base: $a.10^3+b.10^2+c.10+d$

Then, I know that $a,b,c,d$ have to be even, and $a$ equals $0$.

I made a tree as follows:

Then I counted until I found the page that holds the $100$ position, knowing that every green circle holds $25$ pages in it. When I found it I read the previous number because there is no page $0$. That gives me the page with the number $686$.

Then with a similar reasoning the page with the number $888$ holds the $124$ position.

So, I have two questions. First, is this right? Second, is there a better way to approach this problem? I feel there must be a more analytical way to do it.

Answer 1

So the digits you are dealing with are $0,2,4,6,8$ and you are effectively working in base $5$.

So you could write $100_{10}$ in base $5$ and convert it to even decimal digits.

Likewise you can interpret $888_{10}$ as a number in base $5$.

This gives $100_{10}=400_5\equiv 800$ and $888\equiv 444_5 = 124_{10}$

Answer 2

Good approach, but small mistake. Since there is no page $0$, the original page $88$ ends up being the $24$-th position, rather than the $25$-th. Accordingly, page $200$ is on the $25$-th position ... and so page $800$ ends up being in the $100$-th position. But yes, page $888$ ends up on $124$