[rookie1ja]

Easy Deduction - Back to the Number Puzzles
A teacher thinks of two consecutive numbers between 1 and 10 (1 and 10 included). The first student knows one number and the second student knows the second number. The following exchange takes place:
First: I do not know your number.
Second: Neither do I know your number.
First: Now I know.
What are the 4 solutions of this easy number puzzle?

Spoiler for Solution

Easy Savoury - solution
None of the students can have numbers 1 or 10, since they would guess the other one’s number with no problems. I will describe solutions at one end of the interval of numbers 1-10 (the same can be done on the other end).
Information that the second student does not know must be important for the first student. So the first one must expect that the second one has 1 or 3 (if the first one has 2). And as the second student does not know, then he has certainly not 1. So the first pair is 2 and 3.
If the first one had 3, then he would expect the other one to have either 2 or 4. But if the second one had 2 (and the second one would have known that the first one does not have 1), then he would know the number of the first student. However, neither the second student knows the answer – so he has 4. The second pair of numbers is 3 and 4.
Solutions at the other end of interval are 9 and 8 or 8 and 7.

Spoiler for old wording

A teacher thinks of two consecutive numbers between 1 and 10 (Edit: 1 and 10 included). The first student knows one number and the second student knows the second number. The conversation of the students is as follows:
First: I do not know your number.
Second: Neither do I know your number.
First: Now I know.
Will you find all 4 solutions?

[Red Hat]

What happened if the student A has the number five? Student B can have either the number 4 or 6, in which case he could guess that the other student has 3, 5, or 7.

[mrbojangles]

Honestly, when I tried doing this one, I never included 1 or 10 in the first place since the original problem says BETWEEN 1 and 10, meaning the numbers in question are 2 through 9. That might seem nit picky but these ARE brainteasers and it'd be nice if this one was clarified a little bit more. Also, I agree with the other person, what happens in the case of a 5? Why has that one been eliminated?

[earl11]

If it was a 5, then the first one would not be able to say "Now I know". Instead, he would have said "I still don't know". Then, the 2nd person would know that the 1st person has either 4,5, or 6. If the 2nd person had a 3, 4, 6 or a 7, he would be able to say "I know now". But if he had a 5, he would still not be sure. And then the 1st person would know that the 2nd person has a 5.

[courtneycater]

I can think of two situations but not four. If A has the number two, he knows B is either one or three. If B is one, B would KNOW A is two. Since B doesn't know, B must be three. The same would go for eight and nine

I'm getting addicted to these puzzles. I totally forgot to watch TV tonight. Anyone know where to find lateral thinking puzzles?
I guess I could look a little harder but I just can't put off watvching TV any longer

[niilynn]

The questions asks "Will you find all 4 solutions?" The answer, then, should be "yes" or "no." There are a lot of riddles on here that ask that sort of question. The question SHOULD state, "what are the 4 solutions?"

[rookie1ja]

"niilynn":eff69 said:

The questions asks "Will you find all 4 solutions?" The answer, then, should be "yes" or "no." There are a lot of riddles on here that ask that sort of question. The question SHOULD state, "what are the 4 solutions?"

right ... so will you find all 4 solutions? yes or no? there are a lot of people writing the same as you, but just a few of them really answer it (and prove it)

[eedray]

isn't this assuming alot?

[BoilingOil]

I think I CAN see four solutions...

I'll call 1st person A, 2nd person B

A = 2, B = 3: A doesn't know at first. If B had 1, he would know. Since he doesn't know, he must have 3. A solves it at second round
A = 9, B = 8: analogous to the above.

A = 3, B = 4: A doesn't know at first. If B had 2, he would see that A could have only 1 or 3, but with 1, A would have known immediately. Therefor, since B doesn't know, he cannot have 2. A solves it at second round...
A = 8, B = 7: analogous to the above.

I'm still in the process of finding out what happens if A and B were reversed... Maybe there are even more than 4 solutions then

BoilingOil

[DechWerks]

I do not understand why 1 and/or 10 are 'easy' to guess. As this is pretty fundamental, I think this is why I missed this one.