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Old 2020-12-01, 07:19   #1
tgan
 
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Default December 2020

December 2020.

Last fiddled with by LaurV on 2020-12-01 at 07:21 Reason: fixed link
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Old 2020-12-01, 07:41   #2
tgan
 
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Default 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
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Old 2020-12-01, 09:27   #3
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Quote:
Originally Posted by tgan View Post
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
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Old 2020-12-01, 09:58   #4
LaurV
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Yep, you are right on both counts. They will probably correct it later.
I fixed your first post (runaway link).
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Old 2020-12-01, 12:33   #5
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I count 67 unique solutions.

Have puzzles in the past had multiple solutions, or did I just make a mistake somewhere?
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Old 2020-12-01, 14:56   #6
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Quote:
Originally Posted by retina View Post
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
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Old 2020-12-01, 15:24   #7
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Quote:
Originally Posted by retina View Post
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.
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Old 2020-12-01, 15:44   #8
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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
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Old 2020-12-01, 22:28   #9
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Quote:
Originally Posted by Dr Sardonicus View Post
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
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Old 2020-12-02, 00:39   #10
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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
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Old 2020-12-02, 02:44   #11
LaurV
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Quote:
Originally Posted by retina View Post
It all seems a bit muddled
Of course, what did you expect? It is about elections...
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