Question 1046168
Each sheet has two sides, with two consecutive page numbers:
1 and 2, on the first sheet, 3 and 4, on the second sheet, and so on.
The numbers on the {{{n^th}}} sheet were
{{{2n-1}}} on the first side, and
{{{2n}}} on the flip/back side.
Adding both of those numbers, you get {{{(2n-1)+2n=4n-1}}} .
Let's say the {{{43}}} sheets Jerry pulled out were sheets number
{{{n[1]}}} , {{{n[2]}}} , {{{n[3]}}} , ...., and {{{n[43]}}}
The sum of page numbers on those sheets is
{{{(4n[1]-1)+(4n[2]-1) +(4n[3]-1) +"..."+(4n[43]-1) =   4(n[1]+n[2]+n[3]+"..."+n[43])-43=4(n[1]+n[2]+n[3]+"..."+n[43])-44+1=4(n[1]+n[2]+n[3]+"..."+n[43]-11)+1}}} .
The expression {{{(n[1]+n[2]+n[3]+"..."+n[43]-11)}}} is definitely a positive integer,
so {{{4(n[1]+n[2]+n[3]+"..."+n[43]-11)}}} is definitely a multiple of {{{4}}} ,
and the sum,
{{{4(n[1]+n[2]+n[3]+"..."+n[43]-11)+1}}},
could not be a multiple of {{{4}}} .
It could be
{{{4*503+1=2012+1=2013}}} , or
{{{4*504+1=2016+1=2017}}} ,
but it could never be {{{4*504=2016}}} .
(It could not be 2014 or 2015, either).