A nice puzzle I came across. It took some time for me to realize that the question is at all correct!!!The puzzle is as follows:Two brilliant/rational student A and B .They both think of a positive integer(could be more generalized) without knowing the number of the other. They both confides the numbers to the Teacher.They have the following conversations :Teacher:" I am writing sum of the numbers you both have thought and an arbitrary number,say s and u ,Caution there is NO RESPECTIVELY on the board".Teacher to Student A :"Do you know B's number now"Student A:"No"Teacher to Student B :"Do you know A's number now"Student B:"Teacher to Student A :"Do you know A's number now"Student B:"No"..........After finitely many questions and answers..Student A :"YES".How is it possible!!!If possible what is the number of questions required by the Teacher to ask if Student A thinks of the number a and Student B thinks of the number b and number written on the board is c.Assume that both the student can hear each other.Example : Student A thinks of the number 3 and Student B thinks of the number 4 ...on board the Teacher writes 7 and 8 . So Student A thinks B has either thought 4 or 5 which one he doesn't know so he says "NO".Enjoy!!!!

Please neglect this line ..."Student B:"Teacher to Student A :"Do you know A's number now""

"Example : Student A thinks of the number 3 and Student B thinks of the number 4 ...on board the Teacher writes 7 and 8 . So Student A thinks B has either thought 4 or 5 which one he doesn't know so he says "NO"."Not sure I understand the question. If the teacher writes 7 & 8 then A knows that either the teacher added 4 in which case B has 4 or the teacher added 5 in which case B has 2

The teacher writes the sum of their numbers and writes another arbitrary number..But the students doesnt know which one is their sum although they know one of them is their sum of numbers!!!!Another point I want to make it clear that the numbers written on the board remains same through out!!!! I hope you got me!!!

givenstudent A writes astudent B writes bteacher writes smin < smaxproblem: find sum sanswersA1:if a > smin => s = smax qedelse a < sminB1:if b > smin => s = smax qedelse b < sminA2:from B1: a + b < smin + aif smax > smin + a => s = smin qedelse a > smax - sminB2:from A1: a + b < smin + bif smax > smin + b => s = smin qedelse b > smax - sminA3:from B2: a + b > smax - smin + aif smin < smax - smin + a => s = smaxelse a < 2smin - smaxB3:symetric to A3...A4:from B3: a + b < 2smin - smax + aif smax > 2smin - smax + a => sminelse a > 2smax - 2sminbla bla..so... as there are minimal p, q , r ,t in N such thata < p*smax - x*sminb < q*smax - q*smina > smin + r*smin - r*smaxa > smin + t*smin - t*smaxthe problem solved at answer min(p,q,r,t)

the version i saw involves three people, each given a number. then also three numbers on board with one of them is the sum. everyone knows the two numbers of other people but not his own. you can easily extend arkol's proof to this one, or even more general cases.people interested could also google "a headache causing problem" by Conway. there is an interesting anecdote of how Moscow gets Erdos number of 2.----- ----- ----- ----- -----a graph probably explains better:the left panel is for the case of 7 and 8 on board. x and y coordinates are the #s A and B have respectively. each cell with a number represents a possible pair given what are on board, and the number is how many "no"s before you hear "yes".easy to see if A gets 7, there is only one choice: (7,1), so he says yes right away, otherwise there are two choices and he says no;next B's turn: if he gets 7, he says yes, or if he gets 1, he also says yes, as there is also only one choice (6,1) [A would have already said yes for (7,1)], no otherwise;so on so forth...for (3,4), after 6 "no"s, A will finally say yes.the right panel is for the case of 7 and 10 on board. similar reasoning applies.

Hi arkol, your solution is fine apart from bla bla ... coz one would be even interested whose gonna first tell "Yes"!!!Isn't it!!!