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the basement => the archives => Logic Puzzles => Topic started by: Yip on April 29, 2004, 03:55:04 pm
Title: [SPOILER] Varr's Puzzle #5
Post by: Yip on April 29, 2004, 03:55:04 pm
First, if you haven't seen the puzzle, go try it out before reading this.
Now, this time I'm going try something different. I decided allow others to post there solutions and/or comments here. This way I have this space reserved for the "official" answer while hopefully getting more feedback on my puzzles.
Quote
WARNING THIS THREAD MAY CONTAIN THE SOLUTION
Title: [SPOILER] Varr's Puzzle #5
Post by: Katra on April 29, 2004, 09:07:45 pm
Solution follows. Varr will tell whether I got it right.
The key is remembering that pages come in pairs. 1&2 back to back on one sheet and so on.
I figure it at 224 pages left.
Title: [SPOILER] Varr's Puzzle #5
Post by: Yip on May 01, 2004, 04:19:21 am
The key is as you said to remember that the pages come in pairs. This makes the calculations a little tricky. A little trickier than you might think. I think you overlooked something.
In a sense, this puzzle has two solutions since "pages remaining" could mean the pages by number or by actual sheets. I prefer to think of them as sheets myself, though either way 224 is not the answer.
Title: [SPOILER] Varr's Puzzle #5
Post by: Cynic on May 02, 2004, 01:27:36 pm
S p o i l e r
s p a c e . . . ^_^; . . .
I got 288 pages (144 sheets)... Making a program for this was a little harder than I thought; took me a good 15 minutes straight, and it looked like it ran right. 288 seems about right, though.
Title: [SPOILER] Varr's Puzzle #5
Post by: Nocte on May 02, 2004, 02:16:59 pm
I also did it by whipping up a small program. 130 sheets?
Title: [SPOILER] Varr's Puzzle #5
Post by: Katra on May 04, 2004, 11:58:47 am
re-checks math.
124 sheets/248 pages.
Title: [SPOILER] Varr's Puzzle #5
Post by: Lobar on May 04, 2004, 12:47:13 pm
Yet another program returns yet another different answer, here
I got 416 page numbers with a 1 or a 3 in them. and 270 sheets torn out. This is assuming the odd numbered pages are on the right like in most (all?) books. If the odd numbered pages are on the left, only 256 sheets are torn out.
Hope I got it right. It was a fun little program to write.
*edit* Just noticed that the riddle asked for the number of pages remaining, not how many torn out. So there would be 130 sheets left (Unicorn's answer), with the odd numbered pages on the right, and 384 page numbers left in the book. With the odd numbered pages on the left, there'd be 144 sheets left (Cynic's answer) with 289 page numbers left in the book. It's 289 instead of 288 because page 1 would have nothing on the opposite side, so only 1 page number is lost with the sheet.
Title: [SPOILER] Varr's Puzzle #5
Post by: Yip on May 04, 2004, 03:13:13 pm
Writing programs? Hmm, the problem with that approach is that you most likely used the brute force method (otherwise the program would not be necessary). So you got a solution (quite possibly the correct one) but you could hardly show proof here without it being a very long post.
What I'd really like to see is a solution through logical means, not brute force. (or at least not just brute force; you can use it to see if you get the same answer.)
Oh by the way, there are no unnumbered pages at all, not even on only one side. So page 1 is indeed on the right (I'm not counting the inside of the cover as a page)
Title: [SPOILER] Varr's Puzzle #5
Post by: Nocte on May 04, 2004, 03:53:16 pm
Okay, I'm going to try without the aid of a program.
Take 100 pages, or 50 sheets. Per every five sheets, at least 2 will be torn out: the one with pages ..1-..2 (11-12, 21-22,etc), and with pages ..3-..4. That leaves us with 3 * 10 = 30 sheets. The following sheets will also be torn out: 9-10, 15-16, 17-18, 19-20, 29-30, 35-36, 37-38, 39-40.
So we're down to 22 sheets per 100 pages. Now it's easier to look at the entire book:
Title: [SPOILER] Varr's Puzzle #5
Post by: Yip on May 06, 2004, 10:32:36 pm
Here's my solution:
Each sheet has two page numbers (one front and one back). So there are 400 sheets. If we think of it as a three digit number (starting with 001), then the first digit on the remaining pages will be either a 0, 2, or from 4 to 7 (800, the last page number, is on the backside of a sheet), thats six different possible digits. The second digit will be 0, 2, or from 4 to 9, eight different possible digits. and the third digit will be 5&6, 7&8, or 9&0, 3 different possible digits (counting them in pairs). This makes 6 times 8 times 3, or 144 sheets.
But this still includes the sheets where a 1 or 3 appears on the backside but not the front (like sheet 29-30 or 99-100). This happens in one of two ways, either with the first or the second digit (the third digit will never be odd on the backside of the sheet).
For the first digit, its simply 99-100 and 299-300 that still need removed. For the second digit, its the 09-10 and 29-30 for every possible first digit. That is, 6 possible first digits times 2 possible second digits (the third is always a 9 in this case). So 6 times 2 is 12 plus the 2 from the first digit is 14 sheets that still need removed.
144 - 14 = 130 sheets remaining.
Is case you got lost above, here's a breakdown of the remaining sheets: