Question 929300
THE FIFTH-GRADER SOLUTION:
There are {{{9}}} pages with page number < 10,
and it takes only {{{1}}} digit/character to print each page number for those pages,
for a total of {{{9*1=9}}} digits/characters to number those pages.
There are {{{99}}} numbers less than {{{100}}}, so there will be
{{{99}}} pages with one-digit or two-digit page numbers.
Nine {{{(9)}}} of those pages have one-digit numbers, but the remaining
{{{99-9=90}}} of those pages have two-digit page numbers.
As it takes {{{2}}} digits/characters to print the each page number for those pages,
{{{90*2=180}}} digits/characters will be used to number the pages with two-digit numbers.
So far, to number pages 1 through 99, we have used {{{9+180=189}}} digits/characters .
For most books, after page 99, all the other pages have 3-digit numbers,
and it takes {{{3}}} digits/characters to print the each page number for those pages.
If the book has {{{x}}} pages, there will be {{{99}}} pages with one-digit or two-digit page numbers.
If {{{x<1000}}} , the remaining {{{x-99}}} pages will have 3-digit page numbers, and we have {{{927-189=738}}} digits/characters left to print the 3-digit page numbers on those pages.
At 3 digits per page, we can number {{{738/3=246}}} such pages.
After that we will have numbered all {{{9+180+246=highlight(345)}}} pages of the book.
 
TO SHOW OFF YOUR ALGEBRA SKILLS:
{{{9*1+(99-9)*2+(x-99)*3=927}}}
{{{90+90*2+(x-99)*3=927}}}
{{{90+180+(x-99)*3=927}}}
{{{189+(x-99)*3=927}}}
{{{(x-99)*3=927-189}}}
{{{(x-99)*3=738}}}
{{{x-99=738/3}}}
{{{x=246+99}}}
{{{x=highlight(345)}}}