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 2020-12-01, 07:19 #1 tgan   Jul 2015 52 Posts December 2020 Last fiddled with by LaurV on 2020-12-01 at 07:21 Reason: fixed link
 2020-12-01, 07:41 #2 tgan   Jul 2015 52 Posts Error in example? 134 160 206 235 265 = 1000 and not 1001 I think the correct numnber should be 134 161 206 235 265 Last fiddled with by tgan on 2020-12-01 at 07:43
2020-12-01, 09:27   #3
tgan

Jul 2015

52 Posts

Quote:
 Originally Posted by tgan 134 160 206 235 265 = 1000 and not 1001 I think the correct numnber should be 134 161 206 235 265
And [50, 60, 77, 88, 99] only 99 is odd

 2020-12-01, 09:58 #4 LaurV Romulan Interpreter     Jun 2011 Thailand 249916 Posts Yep, you are right on both counts. They will probably correct it later. I fixed your first post (runaway link).
 2020-12-01, 12:33 #5 retina Undefined     "The unspeakable one" Jun 2006 My evil lair 5·1,223 Posts I count 67 unique solutions. Have puzzles in the past had multiple solutions, or did I just make a mistake somewhere?
2020-12-01, 14:56   #6
tgan

Jul 2015

318 Posts

Quote:
 Originally Posted by retina I count 67 unique solutions. Have puzzles in the past had multiple solutions, or did I just make a mistake somewhere?
I think it is possible
must admit that i did not totally understood the puzzle. do we look for a maximum or we need to get the required number?

Last fiddled with by tgan on 2020-12-01 at 14:57 Reason: spelling

2020-12-01, 15:24   #7
EdH

"Ed Hall"
Dec 2009

E6216 Posts

Quote:
 Originally Posted by retina I count 67 unique solutions. Have puzzles in the past had multiple solutions, or did I just make a mistake somewhere?
I remember past puzzles with many answers. A few were "special" and gained extra credit. Special, such that a name may appear or a palindromic answer, etc. within the solutions.

 2020-12-01, 15:44 #8 Dr Sardonicus     Feb 2017 Nowhere 445010 Posts I don't see anything about "the" solution being unique, so why not multiple solutions? I'm not sure exactly what "67 'unique' solutions" means; it seems like an oxymoron. I'm guessing "unique" refers to a specific ordering of the 5 population numbers, so you don't have two solutions with the same five population numbers. Nonuniqueness of solutions is no major defect, but geez -- they can't even give a proper example. Pathetic. Struggling to derive an interesting puzzle from the conditions... It occurred to me to wonder how many ways there are of expressing 1001 as the sum of 5 odd positive integers. Subtracting 1 from each summand gives 5 non-negative even integers. Dividing through by 2, we have the the problem of expressing 498 as the sum of 5 non-negative integers. This is well-known to be equal to the number of ways of expressing 498 as the sum 498 = x1 + 2*x2 + 3*x3 + 4*x4 + 5*x5 where the x's are non-negative integers; the corresponding 5 summands of 498 are x5, x5 + x4, x5 + x4 + x3, x5 + x4 + x3 + x2, and x5 + x4 + x3 + x2 + x1. The number of such 5-tuples is approximately the volume of the 5-dimensional "simplex" in Euclidean 5-space bounded by the coordinate axes and the hyperplane given by the above equation. This volume is 498^5/(5!*5!), which is approximately 2,000,000,000. Last fiddled with by Dr Sardonicus on 2020-12-01 at 16:14 Reason: Forgot extra factor of 5! in denominator
2020-12-01, 22:28   #9
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

17E316 Posts

Quote:
 Originally Posted by Dr Sardonicus I don't see anything about "the" solution being unique, so why not multiple solutions? I'm not sure exactly what "67 'unique' solutions" means; it seems like an oxymoron. I'm guessing "unique" refers to a specific ordering of the 5 population numbers, so you don't have two solutions with the same five population numbers.
If you ignore the ordering, and just consider the raw population counts, then 67 results. If you account for the ordering then a lot more than 67 results.

I only included odd population numbers, as the puzzle says. But the example has even numbers. So I'm not sure what to make of that.

It all seems a bit muddled with the errors and non-conforming examples.

Last fiddled with by retina on 2020-12-01 at 22:29

 2020-12-02, 00:39 #10 Dr Sardonicus     Feb 2017 Nowhere 445010 Posts Upon rereading the problem, I find I hadn't been reading it correctly. I think I have it now, but if so I have a major difficulty with it. A component pi of the vector pop is the numbers of eligible voters in state i. The hypothesis that all the vi are odd insures that there can't be a tie at the ballot box in any state. The corresponding component vi of the vote is the number of votes a candidate got in state i. Therefore vi <= pi, and if 2*vi < pi, then the other candidate gets all that state's electors. OK, the grand total number of electors is given to be 1001. Here is the difficulty I have with that: Both in the example with the non-conforming vector pop and erroneous computation with only 1000 electors being assigned, and in the vector pop in the puzzle itself, the number of electors is greater than the number of eligible voters! Last fiddled with by Dr Sardonicus on 2020-12-02 at 00:41 Reason: xignif posty
2020-12-02, 02:44   #11
LaurV
Romulan Interpreter

Jun 2011
Thailand

100100100110012 Posts

Quote:
 Originally Posted by retina It all seems a bit muddled
Of course, what did you expect? It is about elections...

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