Puzzles with numbers and things

[davis_greatest]

I'll chip in with 9 and 2, although thinking about it I can't see a problem with 8 and 3 either...

Don't think it's 6 and 2 because if it was, Gordon couldn't know that Charlie wouldnt know (as for all Gordon knows, Charlie has the product 15 and therefore would know). If you get me .

I agree that 2 and 6 doesn't work, because that will give sum 8, which means, as you say, that Gordon would think that one possibility for the numbers would be 3 and 5, which if it were the case would give product 15 and then Charlie would have known the numbers. But Gordon knew that Charlie was unable to work out the numbers.

However, it can't be 2 and 9 either. If it were, Gordon would only know that the sum is 11 - he would have no way of knowing that the numbers are not 3 and 8; or 4 and 7.

[davis_greatest]

I agree that 2 and 6 doesn't work, because that will give sum 8, which means, as you say, that Gordon would think that one possibility for the numbers would be 3 and 5, which if it were the case would give product 15 and then Charlie would have known the numbers. But Gordon knew that Charlie was unable to work out the numbers.

However, it can't be 2 and 9 either. If it were, Gordon would only know that the sum is 11 - he would have no way of knowing that the numbers are not 3 and 8; or 4 and 7.

Ditto, robert, it can't be 8 and 3, because Gordon wouldn't be able to be sure that the numbers are not 2 and 9; or 4 and 7.

[Robert602]

Yes you're right of course. I wasn't really considering the last line of the conversation for some reason.

i) Charlie: I couldn't tell the numbers yet.
Gordon: I knew that.
ii) Charlie: Now I can.
iii) Gordon: So can I.

Translated into maths speak:

i) m+n must not be expressible as the sum of two primes. (eg 11, 17, etc)

ii) For all pairs a,b such that a+b=m+n, exactly one pair a,b must satisfy i)

iii) This is the one I'm struggling to get my head around, I'll have another look in the morning, probably.

[snookersfun]

I make it 4 and 13. I don't know how to show a proof though, other than by sending the few columns of Excel that calculated it for me. My spreadsheet shows that that's the only possibility, but I think there are too many pairs of numbers to write something out that is short and elegant by hand to show that nothing else works?

That is the correct answer DG!
If you'll just explain, on what the excel calculations are based, that would do the trick. Obviously, one starts with a long list of possible addition and multiplication results and then narrows it down to one possible result, according to certain criteria.
On second thought, sent me your excel, mine is a mess, would like to see a more elegant method (meaning, do you know how to filter, etc.????)

... and wow, there was a flurry of activity (late at night as well), lets see:

Abextra , 2+6=8, the sum can't be even, as DG, will hopefully explain in his proof.
Robert, some good criteria there....

[davis_greatest]

I think the easiest thing will be if I send you the Excel spreadsheet! You could PM me your e-mail address.

(It won't be for a while though as it's on my home computer - I'm at work now and am not sure if I'll get a chance to turn on my home computer tonight.)

[davis_greatest]

I'll try to explain what the Excel calculations did, as far as I can remember (in simplified terms). I THINK I'm explaining this right, but would need to check with my spreadsheeet.

In columns 1 and 2, set out all possible pairs m, n, such that 1
So the pairs are
(2,3) (2,4) … (2,97) [95 pairs beginning with 2]
(3,4) (3,5) … (2,96) [93 pairs beginning with 3]
…
(j, j+1) (j, j+2) …. (j, 99-j) [99-2j pairs beginning with j]
…
(49,50)

In column 3, give the sum m+n

In column 4, give the product mn


Test 1: "Charlie: I couldn't tell the numbers yet."
For each pair, put an X in column 5 if there is more than one pair with the same product as that pair - X means Test 1 is met

Test 2: "Gordon: I knew that."
For each pair, count in column 6 the number of pairs with the same sum as that pair, for which there is no X in column 5

If there is a zero in column 6, meaning that for all pairs with that sum, there is more than one pair with that product, put an X in column 7 - X means Test 2 is met

If there are Xs in both columns 5 and 7, put an X in column 8 - meaning Tests 1 and 2 are both met

Test 3: "Charlie: Now I can"
For each pair with an X in column 8, count the number of Xs in column 8 for all pairs with the same product as that pair. If the number of Xs is 1, put an X in column 9 - X means Tests 1, 2 and 3 are met

Test 4: Gordon: So can I."
For each pair with an X in column 9, count the number of Xs in column 9 for all pairs with the same sum as that pair. If the number of Xs is 1, put an X in column 10 - X means Tests 1, 2, 3 and 4 are met

Count the number of Xs in column 10. It should be 1 (which it is), showing that there is only one possible solution. The X appears for m=4, n=13

[snookersfun]

That is definitely another point for davis_greatest, who leapfrogged Abextra and Robert into 3rd place, establishing himself as a major contender in this race. (I should stop posting puzzles)
Concerning the explanation: I had hoped for it to cover the mathematical implications a bit (will do some of it a bit later), but the logic is clear.

SO HERE IS THE SCOREBOARD AFTER ROUND 36

snookersfun……………………….…..16
Vidas……………………………………….8½
davis_greatest…………………..……6
abextra……………………………..…...5½
robert602…………………………………5
elvaago...............................1

(some rounds may be worth more than one point)
(especially ones won by davis_greatest)

[snookersfun]

So, as promised, a bit more about the math behind the logic (a note of caution: excuse my English, hope it will not complicate the situation; and please, feel free to correct me, if I blunder somewhere in the process).

Robert started quite well:

i) Charlie: I couldn't tell the numbers yet.
Gordon: I knew that.
ii) Charlie: Now I can.
iii) Gordon: So can I.

Translated into maths speak:

i) m+n must not be expressible as the sum of two primes. (eg 11, 17, etc)

ii) For all pairs a,b such that a+b=m+n, exactly one pair a,b must satisfy i)

iii) This is the one I'm struggling to get my head around, I'll have another look in the morning, probably.

i. means, that Charlie received a number, which has more than one possible unique factorization. Therefore neither the product nor x or y can be prime numbers (nor can the product be of the form prime^2, prime^3 or prime^4).
The fact that Gordon knew that, means, that he received a sum, which, as Robert states, must not be expressible as the sum of two primes. At this point all the even sums get dropped (all possible even-numbered sums can be expressed as the sum of two primes), prime+2, prime+3,….etc.
Btw. also all possible products, which are left with only one possible factorization after realizing f1+f2>99 get dropped here.
At this stage one should be left with the possible sums of 11, 17, 23, 27, 29, 35, 37, 41, 47, and 53.

ii. For all pairs a,b such that a+b=m+n, exactly one pair a,b must satisfy i. Here all the products get dropped, with sums of factors not yielding any of those numbers or more than one of these numbers.

iii. As now the person with the sum knows the answer as well, one has to conclude that his sum appears exactly only once, as the possible result of adding up the factors of a product.

Clarifying the last statement a bit, e.g. 4x7=28, 3x8=24, and 2x9=18 (28, 24, 18 all valid products) all produce a sum of 11 and therefore cannot be the answer, whereas only one eligible product 52=4*13 yields the sum of 17.

[elvaago]

Are you a math professor or something, snookersfun?

[snookersfun]

lol, thanks for the compliment. and still lol did it sound good? DG is the math-wizzard. I am only a measly chemist, trying to remember her math

[elvaago]

This goes way beyond any high school math I've ever done. And I've done 6 years of it and my final grade was an 8.9 (out of 10.)

[snookersfun]

the last puzzle, elvaago? I thought it was more about logic, not much math involved there. DG, what do you think?

[davis_greatest]

...all even numbers can be expressed as the sum of two primes...

Of course, that means even numbers greater than 2. 2 can't be expressed as the sum of 2 primes.

As far as I know, though, that's never been proved, anyway. It is known as Goldbach's Conjecture, whose proof is an unsolved problem dating back over 260 years.

[davis_greatest]

...whereas only one eligible product 51=4*13 yields the sum of 17.

Almost! 52 (!)

[elvaago]

Logic was part of the math curriculum at my first year in University. It was called 'discreet mathematics.' But in reality it was Basic Logic. (I did fail that course twice.)