Book & the Torn Pages

  If total no. of pages in a book is k and the page number is n, then k=n/2
  Sum of all the page numbers = n(n+1)/2 =k(2k+1)
  Last page number = 2k = n. Previous to last page no. =n-1=2k-1
  For k th page, odd page no. is 2k-1 and even page no. is 2k.
  Sum of page no. of k th page= 4k-1
  Tearing of Pages : deduction for 1st page torn 4k1-1 = 3, 2nd page torn 4k2-1=7, 3rd page torn 4k3-1=11, ki th page torn 4ki-1= 3+4(ki-1) .This is ki th term of an AP series with first term 3 and common difference c.d=4.
  Sum of Pages after Single page torn: Hence sum of k pages after ki th page is torn = k(2k+1) -[3+4(ki -1)] = 2k2 +k+1 -4ki
  Sum of Pages after 2 pages are torn:  k(2k+1) -[3+4(ki -1)] - [3+4(kj -1)] = 2k2 +k + 2 - 4(ki+kj)
  Sum of Pages after m pages are torn : 2k2 +k + m - 4(k1+k2+k3 +k4+.... +km ). If the pages are consecutive, ki,kj,kl ..... are consecutive numbers.
  Sum of pages after m consecutive pages are torn : 2k2 +k + m - 4*(1/2)[(km-k0)(km+k0+1)] =2k2 +k + m - 2[(km-k0)(km+k0+1)] where k0 is the page no. before k1.
  Sum of torn pages : S= 4(k1+k2+k3 +k4+.... +km ) - m or  2[(km-k0)(km+k0+1)] - m if the pages are consecutive or

                                 (k1+k2+k3 +k4+.... +km ) = (S +m) / 4 = whole no. and k1,k2,....km each should be < = kmax
  Example: In a book with pages numbering 1 to 100, some pages are torn. The sum of the numbers of remaining pages is 4949. How many pages are torn off ?

Here n=100 and sum of page numbers of untorn book is 100*101/2=5050. Sum of pages left after book is torn is 4949.

Hence sum of torn pages =S=5050-4949=101. (k1+k2+k3 +k4+.... +km ) = (S +m) / 4 = (101+m)/4 . To become whole number, m=3,7,........ If m becomes 7 and supposing that the pages are starting from lowest no. 1, then S=4(1+2+3+.......+7) - 7 =105 which is greater than given 101. So m=3. Hence 3 pages are torn off. k1+k2+k3=26. These 3 pages are not consecutive pages since had these been consecutive pages, their sum sould have been k1+(k1+1)+(k1+2)=3k1+3=26 or k1=23/3 which is not a whole no. Since 1+2+23=26, the maximum page no. cannot be more than 23, otherwise the sum of 3 no. will be more than 26. To find out whether there are 2 consecutive pages, we put that k1+(k1+1)+k2=26 or 2k1+k2+1=26 or k1=(25-k2)/2. Here k2=1,3,5,7,9,11,13,15,17,19,21,23. Out of these,8,9,9 is not feasible. Hence there are 11 arrangements in which 2 pages are  consecutive.

There are 44 ways in which the arrangement can be made so that token total be 26.We have done the chart for 16 tokens which is 28 arrangements. click here. Add for 17->4,18->3,19->3,20->2,21->2,22->1 and 23->1 thus totaling 44.
   

Find no. of torn Pages

Maximum page no. printed in the book (n)
Sum of all *page no. (as printed in the book) (s)
Sum of all page no.(as printed) after few pages are torn (s1)
 
Sum of printed page no. of torn pages (t1)
Minimum no. of pages that are possibly torn (m)
 Torn pages belong to which series ? (series)
To find other possibility of No. of Pages torn including minimum no.-> Put integer values (n1=0,1,2,3.....)
Other possibility of no. of pages torn (m1)(can be changed by puting different values of n1)
Sum of torn page no.(r)
Maximum Page no. of torn page (tmax)
Whether the possibility is feasible ? (fea1)                                                                                                        
   
* Page no. means 1st page, 2nd page, 3rd page etc and page no. as printed means the numbers in print. For 3rd page, page no. is 3 and printed page no. is 5 and 6.