Question 900983
Tom brought a notebook with 100 sheets and numbered its pages consecutively from
1 to 200. Jerry pulled out 43 sheets and added up all 86 page numbers written on
both sides of each of the sheets. can the sum be equal to 2011?
<pre>
Every sheet has an odd number on the front and the next even number on the back.  

Let every front page have odd number {{{2k[i]-1}}} i=1,...,43
and every back page have the next even number {{{2k[i]}}}, i=1,...,43
Thus the 43 sums of the numbers on the 43 pages, front and
back, are {{{4k[i]-1}}}, i=1,...43

We are investigating whether the sum of those 43 sums can be 2011.

Suppose 

{{{sum((4k[i]-1),i=1,43)}}}{{{"="}}}{{{2011}}}

{{{sum((4k[i]),i=1,43)-sum(1,i=1,43)}}}{{{""=""}}}{{{2011}}} 

{{{4sum((k[i]),i=1,43)-43}}}{{{""=""}}}{{{2011}}} 

{{{4sum((k[i]),i=1,43)}}}{{{""=""}}}{{{2054}}}

Divide both sides by 2

{{{2sum((k[i]),i=1,43)}}}{{{""=""}}}{{{1027}}}

Thus we have reached a contradiction since we have
an even number equalling to an odd number.

So we have shown that the sum cannot be 2011.

Edwin</pre>