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tgan 2020-12-01 07:19

December 2020
 
[URL="https://www.research.ibm.com/haifa/ponderthis/challenges/December2020.html"]December 2020.[/URL]

tgan 2020-12-01 07:41

Error in example?
 
134 160 206 235 265 = 1000 and not 1001

I think the correct numnber should be 134 161 206 235 265

tgan 2020-12-01 09:27

[QUOTE=tgan;564911]134 160 206 235 265 = 1000 and not 1001

I think the correct numnber should be 134 161 206 235 265[/QUOTE]

And [50, 60, 77, 88, 99] only 99 is odd

LaurV 2020-12-01 09:58

Yep, you are right on both counts. They will probably correct it later.
I fixed your first post (runaway link).

retina 2020-12-01 12:33

I count 67 unique solutions. :confused:

Have puzzles in the past had multiple solutions, or did I just make a mistake somewhere?

tgan 2020-12-01 14:56

[QUOTE=retina;564929]I count 67 unique solutions. :confused:

Have puzzles in the past had multiple solutions, or did I just make a mistake somewhere?[/QUOTE]

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?

EdH 2020-12-01 15:24

[QUOTE=retina;564929]I count 67 unique solutions. :confused:

Have puzzles in the past had multiple solutions, or did I just make a mistake somewhere?[/QUOTE]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.

Dr Sardonicus 2020-12-01 15:44

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 = x[sub]1[/sub] + 2*x[sub]2[/sub] + 3*x[sub]3[/sub] + 4*x[sub]4[/sub] + 5*x[sub]5[/sub]

where the x's are non-negative integers; the corresponding 5 summands of 498 are

x[sub]5[/sub], x[sub]5[/sub] + x[sub]4[/sub], x[sub]5[/sub] + x[sub]4[/sub] + x[sub]3[/sub], x[sub]5[/sub] + x[sub]4[/sub] + x[sub]3[/sub] + x[sub]2[/sub], and x[sub]5[/sub] + x[sub]4[/sub] + x[sub]3[/sub] + x[sub]2[/sub] + x[sub]1[/sub].

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.

retina 2020-12-01 22:28

[QUOTE=Dr Sardonicus;564939]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.[/QUOTE]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.

Dr Sardonicus 2020-12-02 00:39

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 p[sub]i[/sub] of the vector pop is the numbers of eligible voters in state i. The hypothesis that all the v[sub]i[/sub] are odd insures that there can't be a tie at the ballot box in any state. The corresponding component v[sub]i[/sub] of the vote is the number of votes a candidate got in state i.

Therefore v[sub]i[/sub] <= p[sub]i[/sub], and if 2*v[sub]i[/sub] < p[sub]i[/sub], 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,

[i][b]the number of electors is greater than the number of eligible voters![/b][/i]

:ermm: :confused: :confused2: :shock:

LaurV 2020-12-02 02:44

[QUOTE=retina;564977]It all seems a bit muddled [/QUOTE]
Of course, what did you expect? It is about elections... :razz:


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