SOLUTION: 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

Algebra ->  Permutations -> SOLUTION: 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      Log On


   



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?
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
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?
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%5Bi%5D-1 i=1,...,43
and every back page have the next even number 2k%5Bi%5D, i=1,...,43
Thus the 43 sums of the numbers on the 43 pages, front and
back, are 4k%5Bi%5D-1, i=1,...43

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

Suppose 

sum%28%284k%5Bi%5D-1%29%2Ci=1%2C43%29%22=%222011

sum%28%284k%5Bi%5D%29%2Ci=1%2C43%29-sum%281%2Ci=1%2C43%29%22%22=%22%222011 

4sum%28%28k%5Bi%5D%29%2Ci=1%2C43%29-43%22%22=%22%222011 

4sum%28%28k%5Bi%5D%29%2Ci=1%2C43%29%22%22=%22%222054

Divide both sides by 2

2sum%28%28k%5Bi%5D%29%2Ci=1%2C43%29%22%22=%22%221027

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