Friday puzzle -- pages in a book (corrected)

Re: Friday puzzle -- pages in a book

David Weaver

Hopefully, "any" means non-consecutive.

Because there aren't any consecutive numbers that work out.

My solution is to take pages 1/2 and 3/4 then all of the pages numbered 27-70 (23 sheets of paper, front and back)

And then pick an arbitrary number larger than the 27-70, and find out the average that the sum is short from 2415, and then just add that to the arbitrary pair.

That gives me 135 and 136 for the last pair.

The trick is to pick the middle stretch of numbers to be large enough to make it so the last pair/page you have to pick is smaller than 199/200, but small enough so that the remaining sum isn't from a page already in that middle stretch.

So, anyway, pages 1-4, 27-70 and 135/136

another solution would be, just following the same logic 1-4, 25-68 and 184/185

Re: Friday puzzle -- pages in a book

David Weaver

uh oh, now that the assignment has changed to make the sum 2010, I don't think I can do it.

Each page yields an odd sum.

We want to have 25 pages, so that's an odd sum of odd sums.

The sum of that will be odd. I don't think you can get 2010, but I am lazy today, so I'll sit back and see if anyone else can do it - I could just have a head full of mud.

Correct

Alex Y

Good job on the puzzle as posted. I tried to get cute with the puzzle, and actually made it easier than intended. See subsequent note.

Correct again

Alex Y

Your reasoning is correct. I thought this was too easy, and was trying to get cute with the modification.

Re: Friday puzzle -- pages in a book v3

Alex Y

David knocked the first two out very quickly.

Can you make the pages total to 2413?

Re: Friday puzzle -- pages in a book v3

David Weaver

Yes. Initial solution, but with the final page starting 2 less than the the initial solution.

So, 1-4, 27-70, 134-135.

I'm too lazy to think of a way to make this more difficult, but still possible.

I'm using excel to count and sum the 50 numbers, so I can come up with solutions pretty quickly with the spreadsheet doing the grunt work. I don't know if I can come up with something that won't fit the initial pattern / reasoning for the solution.

( i liked the trivia, though. Enough to make you think, but something you can do reasonably if you have a couple of minutes )

Alex Y

Pages 134 and 135 are not on the same sheet.

Re: 

David Weaver

oh, yeah, good point.

I don't think there is a solution for that. I think the minimum it's possible to increase the sum from any solution is by four.

The reason I think that is because the minimum sum is gotten by pages 1-50

The next smaller sum is gotten by increasing the last two to 51/52

That increases the sum by 4 - from 1275-1279

No matter which numbers are moved in any of the solutions for 2415, they move a minimum of two each, or in at least changing the sum by four.

I could be wrong, I'm not a number theory guy, and that's not exactly a proof, but I think that's the answer.

I could've given you 2411 pretty easily!

My test for whether or not there is a solution would be to take 1275 (the minimum solution) and make sure the difference between that and the target is divisible by four.

(2415 - 1275) = 285 OK

(2413 - 1275) = 284.5 Bad

(

Correct

Alex Y

Another way to look at it is that every sheet has an even number 2*x and its predecessor 2*x-1. So the numbers on any sheet add to a number in the form of 4*x-1. If you add 25 of those, you will get 4*(sum of x's)-25. So the only possible sums of pages on 25 sheets are multiples of four, minus 25.