I have a book with 8 4 pages. One of the pages of the book is torn. If I add the remaining page numbers , the sum is 3 4 5 9 .
Which pages are missing?
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Bonus question: say we didn't know how many pages the book originally had, just that one page had been removed and the sum of the remaining page numbers was 3 4 5 9 . It turns out there's only one additional possible solution; what is it?
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111 = 1 × 111 = 3 × 37
Either it's 1 page which total up to 111 = 55 + 56 or 3 pages with the middle page with sum of 37 = 18 + 19, so the answer is 16~21.
Torn pages
= { 55, 56 } or { 16, 17, 18, 19, 20, 21 }
Is it 1 3 and 1 4 ? In that case there are 8 3 pages in the book? My reason is as follows:
The total numbering on the pages must be an even number and the sum of the page numbers must be more than 3 4 5 9 , in which the total page numbers (i.e. how many pages does the book have) must be greater (or equal) to 8 3 , since the sum 1 + 2 + . . . + 8 3 > 3 4 5 9
The difference between the page numbers and the sum of the remaining page numbers must be odd (since the sum of any 2 consecutive positive integers must be odd). I find that there are only 2 possible solutions in this case, namely 5 5 , 5 6 when page number = 8 4 or 1 3 , 1 4 when page number = 8 3
I don't think there will be any other possible solutions, because the moment you exceed 8 4 , the difference between page numbers and remaining page numbers, expressed as two consecutive positive integers will have exceeded the maximum number of pages in each case, and the difference will only get larger as the number of page numbers increases.
That's all I've worked out for now ^_^ this is a fun extension, thank you!
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That's exactly it! You might just want to prove your third bullet point, but that's why there can't be any other solutions. Thanks for sharing the original problem.
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yeah, I'll (hopefully) prove that tomorrow! Thanks for the extra challenge 👍 have a nice day ahead!
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We first find the sum
1 + 2 + 3 + . . . + 8 3 + 8 4 , which is equal to
8 5 × 4 2 = 3 5 7 0
Since the sum of the remaining page numbers are 3 4 5 9 ,
3 5 7 0 − 3 4 5 9 = 1 1 1
1 1 1 = 5 5 + 5 6 (only possible combination for consecutive numbers)
Therefore Page 5 5 - 5 6 is missing.