I have two seemingly impossible puzzles below:
PUZZLE 1: One page of a book is torn off and the sum of the remaining page numbers is 14001. Can you find the torn page number?
PUZZLE 2: Some pages were torn off from a magazine with page numbers from 1 to 32. If the sum of the remaining page numbers is 495, how many pages were torn off and what are their numbers?
Puzzle two is more tractable but puzzle one has created a lot of confusion and many wrong and misleading statements and solutions have been posted online.
I write this blog to provide proper solutions to these two very interesting mathematical puzzles.
Both puzzles appear to provide too little information for solution.
However, the details are in the way books and magazines are printed and their pages numbered. Following standard practice, I shall call the leaf of a book a page which has two page-faces (two page numbers); I also note that all page-faces are numbered and start from number 1 on page one of the book.
The statements of the puzzles may be supplemented by the following observations:
1. Pages in a book are created by printing four page-faces (2 in the front and 2 in the back) on one sheet of paper and then folding the sheet in half. Books are stitched and bound in a way that the page number of the end page-face is a multiple of 4.
2. The sum of all page numbers is an even number. (refer to point 1)
3. Each page starts with an odd number on the front and has an even number on its reverse side.
4. The sum of page numbers on the two page-faces of a page is an odd number (ref. point 3).
5. All page numbers are integers.
In puzzle 1, one page is torn off the book; Therefore, the sum of remaining numbers must be odd (ref. points 2 and 4)
This has been a problem with several statements of puzzle 1 where the sum of remaining page numbers is given as 10,000 or 15,000 etc - an even number. This is incorrect and makes the puzzle unsolvable.
SOLUTION (PUZZLE 1): One page of a book is torn off and the sum of the remaining page numbers is 14001. Can you find the torn page number?
Also suppose that the page with page numbers P and P+1 has been torn.
Let us work out the sum S of page numbers 1 to N in the book. It is
S = 1 + 2 + 3 + 4 + ........+ (N-1) + N
OR S = N + (N-1) + .................... + 2 + 1
Add the two equations
2S = (N+1) + (N+1) + (N+1) +........ N terms
= N (N+1)
OR S = N (N+1)/2 eq.1
The sum of numbers on the torn page = P + (P+1) = 2P + 1
Therefore, the sum of remaining page numbers, 14001, may be written as
N (N+1)/2 - (2P + 1) = 14001 eq.2
or N^2 + N = 4P + 2 + 28002 = 4P + 28004 eq.3
We note that the minimum value of P is 1 (first page is torn) and the maximum value of P is N-1 (last page is torn).
These limits of P, when used in eq.3, will define the range that N may have.
For P = 1 _____ N^2 + N = 28008
N^2 + N + 0.25 = 28008.25
(N + 0.5)^2 = (167.36)^2
Therefore N = 167.36 - 0.5 = 166.86
For P = N-1_____ N^2 + N = 4N - 4 + 28004
N^2 -3N = 28000
N^2 -3N + 2.25 = 28000 + 2.25 = 28002.25
(N - 1.5)^2 = (167.34)^2
N = 167.34 + 1.5 = 168.84
Therefore, The maximum page number in the book must be in the range 167 to 169. But N also has to be an even number. This gives a unique value of N as 168. We can use this value of N in eq.2 to calculate P - the torn page number.
168 x 169/2 - (2P + 1) = 14001
or 2P = 14196 - 14001 -1 = 194
or P = 97
The answer of puzzle 1 is that the page with page numbers 97 and 98 was torn off. This answer is unique and no other number satisfies the equations.
SOLUTION (PUZZLE 2): Some pages were torn off from a magazine with page numbers from 1 to 32. If the sum of the remaining page numbers is 495, how many pages were torn off and what are their numbers?
Sum of the page numbers (all pages present) of the magazine = 32 x (32+1)/2 = 528
After tearing off some pages, the sum of the remaining page numbers = 495
Sum of the page numbers torn off = 528 - 495 = 33
The sum of the front and back page numbers for a page with front face page number P is P + (P+1) = 2P+1
If an even number of pages are torn then the sum of their page numbers will also be even - this is inconsistent with the odd number 33.
Therefore, an odd number of pages were torn off.
The answer cannot be 1 page torn - then the page numbers will have to be 16 and 17 to add to 33. However on a page, the number on its front page face is odd - not even as in 16.
We can now find actual torn page numbers by trying 3, 5, 7... pages.
For 3 pages, I calculate the following combinations which add to 33.
Front page face numbers 1, 3 and 11 give (1+2) + (3+4) + (11+12) = 3 + 7 + 23 = 33
Front face numbers 1, 5 and 9 give (1+2) + (5+6) + (9+10) = 3 + 11 + 19 = 33
Front face numbers 3, 5 and 7 give (3+4) + (5+6) + (7+8) = 7 + 11 + 15 = 33
For 5 pages, the lowest sum will be if the first five pages were torn. This will give us
(1+2) + (3+4) + (5+6) + (7+8) + (9+10) = 55 which is greater than 33. Therefore 5 or more pages are not possible answers.
The three torn pages have page numbers:
1,3,11 or 1,5,9 or 3,5,7
Note that page numbers 3,5,7 represent consecutive pages. If the puzzle is worded as consecutive pages were torn off the magazine then 3,5,7 becomes a unique answer.
This may be found by the following method.
The page numbers of the three torn pages are
P, P+1, P+2, P+3, P+4 and P+5.
Their sum is 6P+15.
If this is equal to 33 then P = 3 as before.
Hope you have enjoyed the puzzles. Let me know by writing to ektalks@yahoo.co.uk or by leaving your comment on the blog site.
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