Okay, my reckoning is that the numbers are 2 and 2, with a product and sum of 4. Because if either of them has been given 4, they know they can’t solve it as they don’t know if they have the sum (could be 1 + 3, or 2 + 2) or the product (2 x 2). But when Sam says it can’t be solved, then Pru will know that the only combination of both sums and products that does work must be 2 and 2? Basically, the answer lies in a combination that works for one combination of the product that works for two combinations of sums.
Well, I’ve completed the first draft of my logic. I will review it in the next day or so.
I have produced a manuscript spreadsheet (and an Excel one as well) to help illustrate some of the logic.
I need to clarify the text as to how I arrived at 8 for the Sum and 12 for the Product and, just as importantly, how I have eliminated 7 for the Sum (3 + 4) even though 3 x 4 = 12 is an ambiguous product competing with 2 + 6 = 8 and 2 x 6 = 12
But 2 and 6 it is.
Well, to my mind !
Someone put us out of our misery please!
Mike
I will post my solution after breakfast, it is 7.30am here so my brain isn’t quite functioning yet. I think Don is working on an alternative which I am keen to see when it is ready.
Thanks for suggesting a possible solution. Particularly as I had visions of you being distracted as you enjoyed a sundowner with a lovely view, steak and beer.
I think your solution is based on a variation of the problem. For example you allow 1 to potentially be one of the numbers. Also, in my version Sam effectively says “you can’t solve it “. I can’t quite follow what starting conditions you are using. But it is an interesting line of thought.
Mike,
It’s 09:00 and we are about to have breakfast.
The sun is shining. I need to scarify the lawn to get rid of some moss and paint the timber strips that I am going to use to replace the gazebo roof.
No time for a beer or a steak sandwich and that view of the lake and mountains that we enjoy in Canada is 5,000 miles away !!!
But, for you, I wilL post my draft in a couple of minutes and then redraft it later.
This first draft doesn’t quite explain things completely, but it might, hehe, might allow you to drift of to sleep …
… or not !,
Solution Logic
If Pru’s Product is unique, she would have factorized it and identified m and n immediately (m and n are the two elusive numbers, which are not necessarily different).
She didn’t, so the Product of m and n can’t be unique and therefore must be the product of at least 2 distinct pairs of acceptable numbers, AND …
Sam’s statement MUST convey some info that makes it possible for Pru to select the correct pair from amongst those two (or more) different candidates.
That first call from Sam to Pru is key.
Before making that call, Sam will have listed each of the possible pairs of Acceptable Numbers whose Sum is S and noted their corresponding products
These may include Unique Products. But Sam will know that P can’t be one of these because Pru would already have phoned to say “I know your Sum”
Hence, Sam’s list must also contain one or more ambiguous (non-unique) Products, among them P.
However, there is a special case to consider. If there was only one ambiguous Product, Sam would then know that IT had to be P
A solitary, ambiguous product on Sam’s list will always allow Sam to name P
BUT, Sam CAN’T name the Product (AND his phone call to Pru confirms this. He doesn’t say I know your product)
This NOW ENABLES Pru to infer something she didn’t know before.
Pru can now list the POSSIBLE pairs of Acceptable Numbers whose product is P and note their corresponding Sums
Taking each Sum in turn, Pru can now put herself in Sam’s shoes and tabulate what would then be Sam’s candidate Products.
Pru will have inferred that Sam’s actual list must show more than one ambiguous Product.
If it was the situation that one (and only one) of Pru’s candidate Sums gave rise to a list for S showing more than one ambiguous Product, then THAT sum would HAVE to be S.
So, is there a Product that could put Pru in that position?
Well, there is !
it’s 2 + 6 = 8
But I need to add the explanation as to how I figured out that 12 was the Product
and eliminated 3 + 4 = 7 as the sum
But remember, the above is only in draft.
I will post a picture of the spreadsheet I used to help get my thoughts together
Bit scruffy, but it shows the Products and which ones are unique and hence NOT part of the solution etc. It’s only part, remember, the products could go up to 400 (20 x 20) !!
The picture above, could have been drawn by Sam in order to figure out what might be in Pru’s mind.
But Sam has the added knowledge that the Sum is 8 and NOT 7
Only trouble is, WE don’t have that bit of info, (and neither does Pru - obviously)
But because of what Sam said in the first phone call, Pru clearly gets enough info to eliminate the 3 + 4 = 7 option.
BTW I realised I have used “Phone call” rather than “Text message”. I am working from my old notes. Phones were more common when the teaser first emerged !!!
That was all well beyond me I’m afraid.
It’s beyond me, to be honest Mike.
Each time I look at it, I need to modify the explanation. Each time I modify my explanation I run a risk of getting a different answer !!
I know my answer is correct (well, I think it is) because I’ve had it for years.
BTW, I posted the answer on this forum (not this thread) a few days ago
You naughty boy
You probably know where that answer is …
Btw, it’s only the answer, not the explanation.
The Half Crown indeed, well played.
You’re on the ball Mike !
Full marks.
Solution
For this solution I will make an assumption that Sam and Pru were not expected to contact each other as soon as they solved it. Don makes the alternative assumption that they would. I had deliberately left this open - I will explain why in a wrap up once we have all got our heads around this teaser. Assuming we still have the stomach for it of course!
Taking each text in turn we can narrow down the possible number pairings.
TEXT 1:
Sam knows that there isn’t a unique product. Dozey correctly deduced that the numbers cannot both be primes. Hence the sum cannot be formed from two primes. Dozey identified most of the possible sums but Don identified the correct list (up to 40) as 11, 17, 23, 27, 29, 35, 37.
TEXT 2:
Don suggested producing a table of the possible products, with a column for each of the possible sums. In my table below the rows represent the possible values for the lower of the two numbers. We know Pru originally had multiple possible pairings. She can only solve it if only one of them matches the possible sums. So, we can cross out all the products that appear more than once.
TEXT 3:
As Sam can now solve it, he must have only one possible solution. We can see that for the sum of 17, all the products have been crossed out apart from 52. So (4,13) is a solution. None of the other columns have a unique product, so (4,13) is the only solution.
I hope this doesn’t mean that my solution falls down like a house of cards
I’ve been mulling over your 4 + 13 = 17, which is one of the three possibilities that I mentioned earlier in the thread. A couple of things have come to light during this “mulling”
First, you are correct in that I did assume that Pru and Sam are each really trying hard to discover the other’s number (and in turn the two integers) and that they announce (and always would have) their discoveries as soon as they succeed. It never occurred to me to assume otherwise. I don’t think we have any indication as to precisely how clever, or devious either Pru or Sam is, nor how long it takes them to “think”; “add”; “multiply” etc
Secondly, I’ve realised that I made a mistake when I said that the teaser posted here is the same as the teaser I worked on a few years ago. Ok, the telephone v the email was trivial (I think) but on re-reading my ancient teaser, it limited the integers to “…not greater than 20…” Hence the Sum could be 40, but individual numbers m and n (or x and y) were each limited to 20.
This accounts partly for why I kept getting different answers - I kept starting from different premises, and why I sometimes kept ruling out 4 + 13 = 17 (the other factors of 52 are 2 + 26 = 52 and 26 is greater than 20 !!! so Pru would have called out 4 + 13 = 52 right at the start if 52 had been her Product - it’s unique)
There are a couple of other assumptions that I have made, the most important ones being that:
the statements quoted by Sam and Pru are true, and
that both Sam and Pru believe them to be true
Anyway, I’ll see if I can summarize my route to 2 + 6 = 8 etc later on
Cheers
Don
Don
That all fits in with my recollection (plus my more recent research). Not bad for either of us given we are going back over 40 years.
As I recall it the problem developed something like this:
The “Impossible Problem” appeared in Scientific American magazine in 1979. It wasn’t original but it said the limit for the sum was usually 100 but was simplified to each number being no more than 20 to give an overall maximum sum of 40. Those that solved it (as I did) arrived at (4,13) as the unique solution, but then realised that the simplification caused it to be impossible for the reason you have just said. The reaction was to assume the problem had been oversimplified. Removing the 20 limit allowed a solution. The magazine issued a correction a few months later.
Then some years later someone came up with a neater argument. He said why not assume the question is not in error but also possible to solve. He assumed that Sam and Pru would contact the other if they solved it. Previously the question was silent on this. It lead to a unique solution, which I believe is the one that you are heading towards. So if you nail it that would be fantastic!
Not my best piece of logic, but here goes, bearing in mind I have been working on a slightly different version of the Teaser to Ravvie (my numbers are limited to integers from 2 to 20, Ravvie stated in the teaser that the upper bound was a Sum of 40)
Stage 1
Prime numbers and unique Products can be ruled out.
That eliminates product 4, 5, 6, 7, 8, 9, 10 and 11
The next Product is 12, so let’s examine P = 12
If Pru’s Product is 12 then m and n can only be 2 x 6 = 12 or 3 x 4 = 12
And hence Pru will visualize the associated Sum that Sam has, is either 7 or 8
But can’t be certain which, so remains silent
Sam will have also worked this out
Stage 2
But Sam also has a problem
With a Sum of 8, m and n could be 2 + 6 = 8; 3 + 5 =8 or 4 + 4 = 8
3 x 5 As a Product can be ruled out by Sam as it is unique
But that still leaves him with 2 & 6 = §12 or 4 & 4 = §16
It looks like Stalemate
Stage 3
Until Sam says to Pru “ It must be frustrating …”
Pru now knows that Sam can’t have 7
Because that would imply m and n were either 2 + 5 = 10 or 3 + 4 = 12
And since 10 is unique Sam would already have figured out that Pru’s product had to be 12 – which he hasn’t
However, the other consequence of Sam’s remark is that
Pru now knows that Sam’s Sum has to be 8
And declares as such “I have just solved it”
Stage 4
With this new information, Sam can also eliminate 7 and rationalize that his Sum of 8 must be the 2 + 6 = 8 option
But don’t forget, I mis-read the Teaser as posted and have been unwittingly working on a mixture of the two. So don’t accept the above as a solution of the teaser as posed in this thread.
Ravvie’s explanation of his solution is rational, (and for what little it’s worth, I also arrived at 4 + 13 at one stage, only to discard them for some forgotten reason !)