Who will find the number on their own hat first?

Both $A$ and $B$ have numbered hats on their heads. $A$ and $B$ both cannot see his/her own hat, but they can see other one's hat. $A$ sees number on $B$'s hat as $5$ and $B$ sees $A$'s number as $4$. Both $A$ and $B$ have been informed that $A$'s number is multiplication of two positive integers and $B$'s number is addition of those two integers.

First $B$ is asked that whether he/she knows the two numbers on their hats? If the answer is "No", Then the same question is asked from $A$. Likewise this question is repeatedly asked from both $A$ and $B$ until one says "I know". Suppose that both $A$ and $B$ are ideal super genius. Who will first say "I know" and At which question?

• Are the participants aware that their numbers are different from each other, or must they consider the possibility of having the same number? Also, must the two integers be distinct? Feb 19 '16 at 16:55
• Seems like a variation of this question. Feb 19 '16 at 21:57

+-----+                    +-----+
|  4  |                    |  5  |
| mul |                    | add |
--+-----+--                --+-----+--
| ° ° |                    | ° ° |
|  A  |                    |  B  |
\___/                      \___/

First question to B

$B$ sees a $4$ on $A$'s hat which is the product of 2 positive integers. There are 2 possibilities:

• $1*4$ which means the sum on $B$'s hat would be $1+4=5$
• $2*2$ which means the sum on $B$'s hat would be $2+2=4$

There is no unique solution therefore $B$ must answer "No".

First question to A

$A$ sees a $5$ on $B$'s hat which is the sum of 2 positive integers. There are 2 possibilities:

• $1+4$ which means the product on $A$'s hat would be $1*4=4$
• $2+3$ which means the product on $A$'s hat would be $2*3=6$

Again no unique solution, but $A$ is not done yet. He has to check the thoughts of $B$ during his first question for each of the possibilities.

First question to A - Assuming A's number is a 4, what were B's thoughts?

We saw the possibilities for that already, and know there was no unique solution. Therefore $A$ knows that $B$ would answer "No" and must assume that a $4$ on his hat is possible.

First question to A - Assuming A's number is a 6, what were B's thoughts?

Assuming $B$ sees a $6$ on $A$'s hat which is the product of 2 positive integers, there are 2 possibilities again:

• $1*6$ which means the sum on $B$'s hat would be $1+6=7$
• $2*3$ which means the sum on $B$'s hat would be $2+3=5$

Again no unique solution, so $B$ would have to answer "No" and $A$ must assume that a $6$ on his hat is possible as well.

This means that from $A$'s point of view his hat can be either $4$ or $6$. No unique solution, so he must answer "No".

Second question to B

The possibilities for the number on $B$'s hat from $B$'s point of view are still the same: $5$ or $4$. But now he also knows $A$'s answer for his first question, so he has to analyze his thoughts as well.

Second question to B - Assuming B's number is 5, what were A's thoughts?

Again we know that already, and $B$ will come to the same conclusion. That a $5$ on his hat is possible.

Second question to B - Assuming B's number is 4, what were A's thoughts?

Assuming $A$ sees a $4$ on $B$'s hat which is the sum of 2 positive integers, there are 2 possibilities again:

• $1+3$ which means the product on $A$'s hat would be $1*3=3$
• $2+2$ which means the product on $A$'s hat would be $2*2=4$

2 possibilities, but $B$ must also analyze $A$'s thoughts based on $B$'s first answer. If $A$ would assume a $3$ on his hat, there would be only one possibility for a product ($1*3$). Therefore $A$ would know that $B$'s answer would be "Yes" during the first question. As this was not the case $A$ knows there can be no $3$ on his hat. Therefore $A$ would know there is a $4$ on his hat, which is a contradiction, because $A$ answered "No".

Now $B$ knows that there is no $4$ on his hat. There is only one possibility left ($5$) and he can answer "Yes".

B finds his number first.

Explanation:

A knows the two numbers are either 1 & 4 or 2 & 3, because those are the only positive integers that add to 5. So A knows his number is either 4 or 6.
B knows the two numbers are either 1 & 4 or 2 & 2, because those are the only positive integers whose product is 4. So B knows his number is either 5 or 4.
If B saw a 6, he would know that his number was either 7 (6 + 1) or 5 (2 +3).
If A saw 4, he would know that his number was 4 or 3.

So the questioning begins:

1. B is asked if he knows his number. He doesn't, so he says "no."
2. A is asked. A knows that B sees either 6, 4. Neither of those alone gives B enough information to know his number, so A says "no."
3. B is asked. B knows that if A saw a 4, A would know his own number was 3 or 4. But if A's number had been 3, then B could only be 4, and B would have answered right away in the first round. So B knows that A knows that A is not a 3. So B he knows his number is not 4, so it must be 5. "Yes! It is 5!"

• If A saw 4, wouldn't he only know that his number must be 3 or 4? How does he for sure know it's 4. Feb 19 '16 at 17:58
• If his number was 3 (or any prime), B would necessarily know that his own was 4 (or $A+1$) since the only option for the two positive integers added or multiplied are $A$ and $1$.
– Matt
Feb 19 '16 at 18:00
• Hmm. I think I got distracted in the midst of writing my answer. I will edit it. Thank you. Feb 19 '16 at 18:00

Presuming the stipulations I mentioned in my comment above (the two integers are distinct and we cannot have the same number on both hats):

B gets it at the second question. B knows that his number is an addition of two of the factors of 4. This means the only possibilities are 4 and 5. Since we are assuming we can't have the same number on both hats, the only possibility is 5.

EDIT:

Without restrictions:

A gets on on round 3. A knows his possible numbers are 4 and 6. He passes on question #1. B knows his possibilities are 4 and 5. He does not have enough info to eliminate one, so he passes as well. Now, if A's number was 6, B would realize on round 2. In that case, B's possibilities would have been 5 and 7, and had it been 7, A would have got it on round 1 (7 is prime and only has 1 and 7 as factors. A's number could only be 8 in that case), since A did NOT, then the only choice left for B was 5. Since this was not the case, A knows his number is not 6, and therefore can only be 4.

• I'm pretty sure that one of them can figure it out even if integers and hat numbers can be the same. Feb 19 '16 at 17:26