7
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I am thinking of 2 integers $x$ and $y$ which follow the following; $$2≤x≤y$$ $$x+y≤47$$

I then write the sum $S = x+y$ and the product $P=xy$ on two separate cards.

I pass the sum card to player A and the product card to player B. Neither player knows what is on the other person's card or what the numbers $x, y$ are.

Player A then says "I know that you cannot work out $S$".

Player B then says "I have learnt something but still don't know $x$ and $y$".

Player A then says "I have learnt something and I do know $x$ and $y$".

Player B then says "Finally I know $x$ and $y$ as well".

What must the unit digit of $S$ be?

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From Player A's first statement, we know that

S must be odd.

If S were even, then there's always the possibility that x and y could both be prime. In that case, P would a product of two primes, which would give away x and y for B, contradicting A's first statement that such a thing must be impossible.

By the exact same token,

Any odd S which is two larger than a prime is impossible.

In addition to that,

There are some number pairs (like x=10, y=25), where you have composite numbers on one or both sides, but given the restriction that x+y < 47, the factorisation of the product is unique.

These are the sums that I found that weren't excluded by the previous two steps. Each is listed with one possible pair of x and y for which the product would give away the factors: 27 (4, 23), 29 (6, 23), 35 (4, 31), 37 (6, 31), 41 (4, 37), 47 (4, 43).

Here are the disqualifying reasons for all the possible sums S, given as the number of the spoiler block that disqualifies them:

    Ones:| 0 1 2 3 4 5 6 7 8 9
 --------+-----------------------
 Tens: 0 | - - - - 1 2 1 2 1 2
       1 | 1   1 2 1 2 1   1 2 
       2 | 1 2 1   1 2 1 3 1 3 
       3 | 1 2 1 2 1 3 1 3 1 2 
       4 | 1 3 1 2 1 2 1 3 - -
 

So B knows that A's sum must be one of the remaining numbers:

11, 17, or 23.

From B's first statement, we know that

B's product must be expressible as a product of two numbers in (at least) two different ways, so that one way corresponds to one possible sum, and the other corresponds to another. If that weren't the case, then B would know x and y at this point already.

By listing all the possible number pairs (of which there are 21 at this point), and cross-referencing the products against the sums, we find that B's product must be one of the following:

30: (5,6, sum 11) or (2,15, sum 17)
42: (3,14, sum 17) or (2,21, sum 23)
60: (5,12, sum 17) or (3,20, sum 23)

Out of these possibilities, the only product, where knowing the sum would help A know x and y is

the first one, particularly the one with x=5, y=6, because if the sum were either 17 or 23, there would still be several choices, and A's second statement would be false.

Therefore, even before B's final statement (which naturally does fit, since even we now know the numbers), we have deduced that

x=5, y=6, S=11, P=30

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  • $\begingroup$ Beat me to it. Way to go. $\endgroup$ – Dylan Dec 17 '18 at 14:56
  • $\begingroup$ I'm not sure I follow your first assumption. ROT13: Ubj qbrf gur snpg gung F vf rira tvir njnl k naq l? Vs F jrer 14, sbe rknzcyr, k naq l jbhyqa'g unir gb or 7 naq 7 (gjb cevzrf); gurl pbhyq or 2 naq 12, 4 naq 10, rgp. Yrg nybar gung 14 pna or gur cebqhpg bs qvssrerag cnvef bs cevzrf (r.t., 3+11). $\endgroup$ – Ahmed Abdelhameed Dec 17 '18 at 15:10
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    $\begingroup$ @AhmedAbdelhameed ROT13: Cynlre N xabjf O pnaabg xabj gur fhz. Urapr zhfg or bqq. Vs fhz jnf rira naq O unq n cebqhpg bs 2 cevzrf ur jbhyq npghnyyl xabj gur fhz jvgubhg N fnlvat nalguvat. Fb sbe svefg fgngrzrag gb or gehr gura fhz vf bqq. $\endgroup$ – Ben Franks Dec 17 '18 at 15:20
  • $\begingroup$ Just in case anyone wonders how to make sense of the requirement to only answer with the final digit, assuming that A and B don't know of the <47 limitation doesn't change the answer, it only creates an order of magnitude more cases to check. (Source: just spent more than an hour figuring that out.) $\endgroup$ – Bass Dec 17 '18 at 22:43

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