Question 925269
In all books I have ever had, left-hand pages contained even page #s and the right-hand pages contained odd page #s. That sounds normal to me.
 
If one page is missing, the page on the reverse must be missing too.
For example, if page 1 is missing, page 2, which is the other side must also be missing.
So, missing pages must come in pairs: one odd numbered (right page), with number {{{N}}} ,
and the even numbered page with the next page number, {{{N+1}}} .
If just one such set of 2 pages is missing, the sum of their numbers will be
{{{N  + (N+1) = 2N + 1}}} .
 
The numbers on all 50 pages add up to
{{{(1+50)*50/2=1275}}}
The sum of numbers on the missing pages will be
{{{1275-1242=33}}}
Could that be just 2 pages, an {{{odd}}} page and the next page?
I don't think so, and I can explain why it cannot be so.
If the pages missing were page {{{N}}}, and page {{{N+1}}} ,
{{{2N+1=33}}}-->{{{2N=33-1}}}-->{{{2N=32}}}-->{{{N=16}}} and that does not work, because {{{16}}} is not odd.
The book cannot be missing just page 16 and page 17, because the reverse of page 16 is page 15 and the reverse of page 17 is page 18.
 
Since the sum of the numbers of the missing pages is {{{33}}} which is odd,
and it cannot be the sum of the numbers of 1 pair of consecutive pages,
it must be the sum of the numbers of 3 pairs of consecutive pages,
because the sum of 2 pairs of consecutive pages would be even.
Making the numbers in 2 of the 3 pairs as small as possible would make the numbers in the third pair as large as possible.
The smallest 4 numbers are 1, 2, 3, and 4, and
{{{1+2+3+4=10}}}
The sum of the third pair would be {{{33-10=23}}} ,
and {{{2N+1=23}}}-->{{{2N=23-1}}}-->{{{2N=22}}}-->{{{N=11}}} gives us an odd first page.
The reverse has the number {{{N+1=highlight(12)}}}
and that would be the greatest page number that could be on a page missing from Drew's book.