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Old 2019-01-01, 23:33   #23
SmartMersenne
 
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Quote:
Originally Posted by a1call View Post
It is quite an interesting challenge.
I have tens of glasses which are 15/16 full but I am quite peptomestic about filling the last 1/16 of any of them.
One interesting aspect is that the square root of the sums seem to form in interesting and different patterns.
Many are symmetrical around a 45° diagonal.
Others form a row of primes then primesx2 then semi-primes then semi-primesx2 and other patterns as well.
And of course my posts would not be complete without a few painfully annoying nags to some:
The challenge does not define any rules regarding repeating terms within/between the two sets A & B. Without such clarifications many trivial solutions can be found.
Yup: A=[3, 99, 3, 99] B=[1, 22, 1, 22]
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Old 2019-01-02, 00:46   #24
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Originally Posted by SmartMersenne View Post
Yup: A=[3, 99, 3, 99] B=[1, 22, 1, 22]
AFAIK there is no such thing as repetition of elements in a set. A set is not the same thing as an ordered tuple.

The two sets need not be disjoint, however. Clearly, adding a number to all the elements of one set, and subtracting it from all the elements of the other, creates two sets giving the same sums. So instead of [1,22] and [3,99] we could take [2,23] and [2,98].

Last fiddled with by Dr Sardonicus on 2019-01-02 at 00:47 Reason: gixfin posty
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Old 2019-01-02, 00:58   #25
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Quote:
In mathematics, a set is a collection of distinct objects,
https://en.wikipedia.org/wiki/Set_(mathematics)

Thank you for the correction.
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Old 2019-01-02, 01:08   #26
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Quote:
Originally Posted by Dr Sardonicus View Post
AFAIK there is no such thing as repetition of elements in a set. A set is not the same thing as an ordered tuple.

The two sets need not be disjoint, however. Clearly, adding a number to all the elements of one set, and subtracting it from all the elements of the other, creates two sets giving the same sums. So instead of [1,22] and [3,99] we could take [2,23] and [2,98].
multiset is the term for an object like a set, but with multiplicities. the sums need all be squares, and since squares are 0 or 1 mod 4 , any term in A that is 2 mod 4 forces the elements of B to all be 2 or 3 mod 4. presence of an element that is 1 mod 4 forces the opposite set to all be 0 or 3 mod 4. 3 mod 4 forces all elements of the opposite set to be 1 or 2 mod 4. and 0 mod 4 forces 0 or 1 mod 4. mod 9 all squares are 0,1,4, or 7. etc. anyways time to use forpart ( or setbinop etc doh )and forsubset to try my hand at coding it.

Last fiddled with by science_man_88 on 2019-01-02 at 01:13
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Old 2019-01-03, 18:58   #27
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Default triplet with a pair

Until now, i only have a triplet with a pair :

A = {22, 97, 526}; B = {3, 99};

22 + 3 = 25 Factor {{5^2}}

22 + 99 = 121 Factor {{11^2}}

97 + 3 = 100 Factor {{2^2},{5^2}}

97 + 99 = 196 Factor {{2^2},{7^2}}

526 + 3 = 529 Factor {{23^2}}

526 + 99 = 625 Factor {{5^4}}
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Old 2019-01-03, 19:31   #28
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Default it is going well

it is going well: quadriplet with pair:

B = {3, 99, 4803, 45699} ; A = {97, 526};
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Old 2019-01-04, 05:57   #29
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Originally Posted by a1call View Post
What exactly are you referring to when you say false?
Please be more specific. Thanks.
He refers to the fact that if you don't have the restriction of distinct primes, then Alice (the first choicer) wins every time by dividing with the number itself.

25 -> 1
100 -> 1
etc.

This problem has not much to program, is pure math, and quite simple actually. As said above, N must always be a perfect square. The sums can not be square free, because Alice would win in one shot, picking the whole number as the first divisor, and they can't be non-squares, because then Alice will chose the first time in such a way to leave a perfect square, and she wins. For example, if \(N=2^a3^b5^c...\) then Alice picks first time the product of all primes which have the odd power, and then what is left is a perfect square. Now, the only left for you is to prove that if the number is perfect square, then Bob (the second picker) wins every time. There is a theorem which states that the only numbers with an odd number of divisors are the perfect squares (why? and why do you need an odd number of divisors?) anyhow this is not complicate to prove, but you don't need to go so deep, because there is a simple strategy to win: if N is perfect square, then any prime in N has its pair, and all Bob has to do is to pick the same product Alice picks, every time, leaving every time behind a perfect square. Start with a perfect square, end with a smaller perfect square, repeat. This ends in 1, and it is the "infinite descent" method, invented by Fermat, hehe - as members of mersenneforum, you should know that :razz:

So, all the fuss is about finding two sets of numbers whose paired sums are always perfect squares. How difficult is that? (from the number of the persons who already solved the puzzle in just few hours, and from this current thread, you can see that the puzzle is trivial - so this is another skip this month).

Last fiddled with by LaurV on 2019-01-04 at 06:32
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Old 2019-01-04, 06:38   #30
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And how does any of that have to do with the text that he quoted from me and labeled as false?
Where is exactly the error in my post?
Please make a direct reference to the false statement.
Thanks.
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Old 2019-01-04, 07:17   #31
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Quote:
Originally Posted by a1call View Post
Are primes products of distinct primes?
I would say not, but know of Wikipedia worshippers which would disagree.
Your posts are generally just bullshit, insinuations, insulting the intelligence of the audience, and blah blah blah, so I do not read them in detail anymore, for long time. Generally, primes are primes, can not be composite (i.e. product of primes), and well, even if you reformulated in the next post, to use Title case, which makes more sense (as the "set" of divisors, but is is still extremely odd to call them Primes, as they are mostly composite). However, we can satisfy you this time, with the exact stupidity that you say:

Quote:
Originally Posted by a1call View Post
Else if Primes are-not products of distinct primes then:
4, 25 and 121 have no winners
Oh? How come? I just said how Alice wins in this case...

Last fiddled with by LaurV on 2019-01-04 at 07:18
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Old 2019-01-04, 07:42   #32
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Originally Posted by LaurV View Post
Oh? How come? I just said how Alice wins in this case...
As I understand it, the sticking point for a1call is the "product of distinct primes" bit. According to him, a single prime is not a product -- there should be at least two.

EDIT:- Looks like they have updated "divides N by any divisor that is either a prime or a product of distinct prime numbers". That should take care of that.

Last fiddled with by axn on 2019-01-04 at 07:53
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Old 2019-01-04, 09:49   #33
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Quote:
Originally Posted by axn View Post
EDIT:- Looks like they have updated "divides N by any divisor that is either a prime or a product of distinct prime numbers". That should take care of that.
yeah I contacted the puzzlemaster and posted the response in this thread.
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