538 Puzzle 4/1/16

[silverscreenselect]

Here's this week's 538 Puzzle, and it's a logical puzzle made somewhat simpler by the fact that we're dealing with whole numbers here and not an entire range of numbers:

Three very skilled logicians are sitting around a table — Barack, Pete and Susan. Barack says: “I’m thinking of two natural numbers between 1 and 9, inclusive. I’ve written the product of these two numbers on this paper that I’m giving to you, Pete. I’ve written the sum of the two numbers on this paper that I’m giving to you, Susan. Now Pete, looking at your paper, do you know which numbers I’m thinking of?”

Pete looks at his paper and says: “No, I don’t.”

Barack turns to Susan and asks: “Susan, do you know which numbers I’m thinking of?” Susan looks at her paper and says: “No.”

Barack turns back to Pete and asks: “How about now? Do you know?”

“No, I still don’t,” Pete says.

Barack keeps going back and forth, and when he asks Pete for the fifth time, Pete says: “Yes, now I know!”

First, what are the two numbers? Second, if Pete had said no the fifth time, would Susan have said yes or no at her fifth turn?

They also gave the answer to last week's puzzle, and, although most of the people here figured out it was a good bet for the player, no one figured out just how good a bet it was. And the answer turns out to be fairly simple. It's $100 x Euler's number (2.71+). The explanation of how they arrived at that, along with this week's puzzle, can be found here.

http://fivethirtyeight.com/features/can ... le-puzzle/

[jarnon]

Spoiler

2 & 8. 2 x 8 = 16; no other remaining combination has the product 16 (4 & 4 was eliminated on Susan's last turn).

Susan would say no. 2 & 8 would be eliminated, but there are still two pairs with sum 10: 4 & 6, and 1 & 9. The game could go on forever this way.

[BackInTex]

Barack must be able to choose the same number in a pair. Not choosing the same number, Pete should know the numbers on his 3rd ask.

I'll go back and work the problem assuming the pair could be the same number and get back to you.

[BackInTex]

First, what are the two numbers?

4 and 4

Second, if Pete had said no the fifth time, would Susan have said yes or no at her fifth turn?

Hmmm. Unless the answer is "Maybe" I may have done something wrong. If the numbers weren't 4 and 4 I'm showing the remaining pairs after Pete's 5th "no" to be:

1 & 8
2 & 4

1 & 9
3 & 3

2 & 8

2 & 9
3 & 6

3 & 8
4 & 6

resulting in a "Yes" if the numbers were 2 & 8 and "No" for the others

[BackInTex]

Spoiler

2 & 8. 2 x 8 = 16; no other remaining combination has the product 16 (4 & 4 were eliminated on Susan's last turn).

Susan would say no. 2 & 8 would be eliminated, but there are still two pairs with sum 10: 4 & 6, and 1 & 9. The game could go on forever this way.

But 4 & 6 and 1 and 9 don't have the same product, so while Pete would say "no", Susan would have either 24 or 9 on her paper and say "yes".

But I did not get to where you ended. See my response above. I missed something somewhere.

[plasticene]

Spoiler

2 & 8. 2 x 8 = 16; no other remaining combination has the product 16 (4 & 4 was eliminated on Susan's last turn).

Susan would say no. 2 & 8 would be eliminated, but there are still two pairs with sum 10: 4 & 6, and 1 & 9. The game could go on forever this way.

I completely agree. I mis-remembered what the question was asking, so I initially wrote something stupid.

[plasticene]

First, what are the two numbers?

4 and 4

Second, if Pete had said no the fifth time, would Susan have said yes or no at her fifth turn?

Hmmm. Unless the answer is "Maybe" I may have done something wrong. If the numbers weren't 4 and 4 I'm showing the remaining pairs after Pete's 5th "no" to be:

1 & 8
2 & 4

1 & 9
3 & 3

2 & 8

2 & 9
3 & 6

3 & 8
4 & 6

resulting in a "Yes" if the numbers were 2 & 8 and "No" for the others

You're almost there, except the numbers can't be 4 & 4, because it was the only pair left that added up to 8 when Susan gave her fourth response, so she would have known 4 & 4 was the answer. (And if 4 and 4 were somehow still around for Pete's fifth response, he couldn't say he knew, because both 4 & 4 and 2 & 8 have the same product.)

Your list of pairs is the complete list of possibilities after Susan's fourth NO. Since Pete is now sure which one is right, the only possibility is 2 & 8. If Pete said NO on his fourth response, Susan could eliminate 2 & 8, but none of the rest of the candidates, since there would be two pairs adding up to 6, two adding up to 9, two adding up to 10, and two adding up to 11.

[Bob78164]

First, what are the two numbers?

4 and 4

Second, if Pete had said no the fifth time, would Susan have said yes or no at her fifth turn?

Hmmm. Unless the answer is "Maybe" I may have done something wrong. If the numbers weren't 4 and 4 I'm showing the remaining pairs after Pete's 5th "no" to be:

1 & 8
2 & 4

1 & 9
3 & 3

2 & 8

2 & 9
3 & 6

3 & 8
4 & 6

resulting in a "Yes" if the numbers were 2 & 8 and "No" for the others

You're almost there, except the numbers can't be 4 & 4, because it was the only pair left that added up to 8 when Susan gave her fourth response, so she would have known 4 & 4 was the answer. (And if 4 and 4 were somehow still around for Pete's fifth response, he couldn't say he knew, because both 4 & 4 and 2 & 8 have the same product.)

Your list of pairs is the complete list of possibilities after Susan's fourth NO. Since Pete is now sure which one is right, the only possibility is 2 & 8. If Pete said NO on his fourth response, Susan could eliminate 2 & 8, but none of the rest of the candidates, since there would be two pairs adding up to 6, two adding up to 9, two adding up to 10, and two adding up to 11.

That's the answer I get as well: 2 and 8. --Bob

[BackInTex]

Found my error. I had Pete having the sum and Susan the product. Doh! "Read the problem twice!"