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I had three mathematician friends at school, and we were hanging out.
Out of the blue a question came to my mind, and I decided to ask it to them:

"Guys, I have a question for you. Which is actually pretty easy, I am sure you will find it very fast! I will think of 2 numbers which are distinct and they are between 1 and 10 (including 1 and 10).

I will tell the product of them to you Matt,

and I will tell the absolute difference of them to David,

and the sum of them to you Simon.

They started to talk each other:

Matt: I cannot find it guys.

David: I cannot find it either.

Simon: This is very hard question actually, I do not believe anyone will ever find it.

Matt: I cannot believe that I cannot find it still, these are pretty good chosen numbers.

Simon: You made a very hard question indeed. Still could not figure it out.

David: I found it! I thought I would never find it, thanks Simon!

Matt: We all found it then.

Simon: Yes. Good one thanks.

What were those numbers?

Note: This is not a duplicate question; it is a unique one. Just inspired from this one. The logic and solution are totally different. If you still believe that it is the same one, I am okay anyway.

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6
  • 6
    $\begingroup$ It is not a duplicate question, just a question which is the same type. $\endgroup$
    – Oray
    Jan 25, 2016 at 20:46
  • 1
    $\begingroup$ @JoeBeastlyGerbil Then puzzling.stackexchange.com/questions/251/… is the original one and 3 impossibly intelligent mathematicians were a duplicate too? $\endgroup$
    – Oray
    Jan 25, 2016 at 20:51
  • $\begingroup$ Depends which came first, and if the user new of the other question and also if both of the posters new about the other question $\endgroup$
    – Beastly Gerbil
    Jan 25, 2016 at 20:56
  • $\begingroup$ Inspired puzzles have to have a different solution and question $\endgroup$
    – Beastly Gerbil
    Jan 25, 2016 at 20:57
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    $\begingroup$ This is not a duplicate. This problem has a much smaller search space (1 to 10 versus 1 to 1000), and the conversations the three people have are different. Also, the other problem could only be solved by computed, while this is tractable by hand. This is a variant on a theme, which is usually encouraged on PSE. $\endgroup$
    – Mike Earnest
    Jan 25, 2016 at 21:47

3 Answers 3

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11
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The solution is:

2 & 9 - product of 18, absolute difference of 7, sum of 11.

First we can eliminate all pairs of unique numbers (1-10 inclusive) that would give unique products. Since Matt doesn't know what the pair of numbers is, there must be more than one way to get the product he was given.

Thus, we eliminate the following products: 2, 3, 4, 5, 7, 9, 14, 15, 16, 21, 27, 28, 32, 35, 36, 42, 45, 48, 50, 54, 56, 60, 63, 70, 72, 80, and 90.

First elimination by Matt

Next, David, having eliminated all pairs that give unique products checks to see which pairs yield the absolute difference he was given. Since he does not know the pair at this point, we can assume that there is more than one pair yielding the absolute difference.

We therefore eliminate the unique absolute differences: 4, 6, 8, and 9.

Now we come to Simon, who is unable to determine the pair by the sum he was given. We can therefore assume that, of the pairs remaining, there is more than one pair of numbers that add up to that sum.

So we eliminate the unique sums: 5, 6, and 10.

Having seen all of these eliminations, Matt still does not know the answer. This means that the product is still not unique, so we will eliminate the unique products:

6, 8, 10, 12, 20, 24, and 40. This leaves only 4 pairs remaining.

With 4 pairs remaining, Simon still doesn't know the answer, which means that the sum he was given is still not unique. We now eliminate the unique sums:

9 and 13. We only have 2 pairs left, both of which have a sum of 11.

David was still torn (between pairs with a non-unique absolute difference) until Simon spoke up. The non-unique pairs he was seeing were:

2 & 9 (difference of 7) and 3 & 10 (difference of 7).

But only ONE of these two pairs remains after Simon's elimination. David therefore knows that the correct pair is:

2 & 9. enter image description here

Matt and Simon, understanding the conclusion David has drawn, also agree.

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  • $\begingroup$ yes you did it right, i did a small mistake there, will fix it and u can edit accordingly. $\endgroup$
    – Oray
    Jan 25, 2016 at 22:08
  • $\begingroup$ I fixed the question to create only one answer. try to change the last part and find it please :) $\endgroup$
    – Oray
    Jan 25, 2016 at 22:32
  • $\begingroup$ @Oray, with your most recent edit, I am now unable to find a solution. I have two options for how Matt and David could solve it, but I don't know how Simon will have a different response between his second and third statements. $\endgroup$
    – tinezekis
    Jan 25, 2016 at 22:42
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    $\begingroup$ Well, David found it out because Simon could not find it, otherwise David would not able to find it. That's why he thanks to Simon, means if David responded before Simon, he would not able to figure it out. $\endgroup$
    – Oray
    Jan 25, 2016 at 22:51
  • $\begingroup$ Gotcha, @Oray. I have edited my original solution. Is this what you got as well? $\endgroup$
    – tinezekis
    Jan 26, 2016 at 3:15
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9 AND 2 because 5,6 would have been a duplicate, had Simon not been confused first Conditional Formatting

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2
  • $\begingroup$ Welcome to Puzzling.SE! Can you explain your spreadsheet? $\endgroup$
    – Deusovi ♦
    Jan 26, 2016 at 7:56
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    $\begingroup$ Well, on the first run, M1 (Matt 1) I marked the duplicates of multiples, with orange, then on the second D1, I marked the duplicates falling under the duplicates of the previous even darker and so on and so forth... $\endgroup$
    – Ternary_lover
    Jan 26, 2016 at 7:59
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Here's my solution:

The numbers are 5 and 6.

First elimination:

Since David doesn't know the number, the product must be unique. This leaves us with the following pairs: 1-6, 1-8, 1-10, 2-3, 2-4, 2-5, 2-6, 2-9, 2-10, 3-4, 3-6, 3-8, 3-10, 4-5, 4-6, 4-10, 5-6, 5-8.

Second elimination:

We know that the absolute difference is not unique, so we eliminate those pairs with differences of 4, 6, 8 and 9. That leaves us with (organized by increasing difference): 2-3, 3-4, 4-5, 5-6, 2-4, 4-6, 2-5, 3-6, 5-8, 1-6, 3-8, 1-8, 2-9, 3-10

Third elimination:

The sum is still not unique, so eliminate those with sums of 5, 6, or 10. This leaves us with (in order of increasing sum): 3-4, 2-5, 1-6, 4-5, 3-6, 1-8, 5-6, 3-8, 2-9, 5-8, 3-10.

Fourth Elimination:

The product is STILL not unique, which narrows this down to just four pairs of numbers. We have either a product of 18 with 2-9 and 3-6 or a product of 30 with 5-6 or 3-10.

Fifth Elimination:

The sum is STILL not unique, which means we must have either 2-9 or 5-6.

Sixth Elimination:

David was able to get it thanks to Simon's last statement, which means that previously (in the 2-9, 3-6, 5-6 or 3-10 step) the difference was not unique. Since it is now, we can eliminate 2-9 (since 3-10 has the same difference). The numbers are 5-6.

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