@gyrodceite said:There are 5 thieves (Different age )with 100 Diamonds.Oldest thieve among 5 will propose a distribution of diamonds.If 50% of the votes( do not take oldest thief in account) are in favor then oldest one will survive otherwise rest of the thieves will kill the oldest one.All the thieves are rational thinkers and good at maths.Nobody wants to die.So you are the oldest thief and come out with a nice proposal so that you can survive and take away the maximum share..
@gyrodceite said:There are 5 thieves (Different age )with 100 Diamonds.Oldest thieve among 5 will propose a distribution of diamonds.If 50% of the votes( do not take oldest thief in account) are in favor then oldest one will survive otherwise rest of the thieves will kill the oldest one.All the thieves are rational thinkers and good at maths.Nobody wants to die.So you are the oldest thief and come out with a nice proposal so that you can survive and take away themaximum share..
@gyrodceite said:There are 5 thieves (Different age )with 100 Diamonds.Oldest thieve among 5 will propose a distribution of diamonds.If 50% of the votes( do not take oldest thief in account) are in favor then oldest one will survive otherwise rest of the thieves will kill the oldest one.All the thieves are rational thinkers and good at maths.Nobody wants to die.So you are the oldest thief and come out with a nice proposal so that you can survive and take away the maximum share..
@bs0409 said:ANS=97The trick is to work backwards.Lets assume A>B>C>D>E are the thieves with A being seniormost.If only D and E are left, then D will die anyway.(Because if D dies, E gets all the 100 diamonds)Given this, if C,D,E are left, D will vote in favour of C and E will oppose C[since D does not want to goto the last round and E wants to go there]So,C will win the round anyway because of Tie.So C offers 100 for himself,0 for D,E.When B,C,D,E are left, C wants to goto next round.For D,E it doesn't matter. For B, he wants to win here only.So he offers B:98,C:0,D:1,E:1 to get votes of D and ESimilarly A will offer A:97,B:0,C:1,D:1,E:1
@bs0409 said:ANS=97The trick is to work backwards.Lets assume A>B>C>D>E are the thieves with A being seniormost.If only D and E are left, then D will die anyway.(Because if D dies, E gets all the 100 diamonds)Given this, if C,D,E are left, D will vote in favour of C and E will oppose C[since D does not want to goto the last round and E wants to go there]So,C will win the round anyway because of Tie.So C offers 100 for himself,0 for D,E.When B,C,D,E are left, C wants to goto next round.For D,E it doesn't matter. For B, he wants to win here only.So he offers B:98,C:0,D:1,E:1 to get votes of D and ESimilarly A will offer A:97,B:0,C:1,D:1,E:1
@gyrodceite said:There are 5 thieves (Different age )with 100 Diamonds.Oldest thieve among 5 will propose a distribution of diamonds.If 50% of the votes( do not take oldest thief in account) are in favor then oldest one will survive otherwise rest of the thieves will kill the oldest one.All the thieves are rational thinkers and good at maths.Nobody wants to die.So you are the oldest thief and come out with a nice proposal so that you can survive and take away the maximum share..
@sails said:how can you be sure that C,D,E will be satisfied with 1 diamond..if they are intelligent then they should settle for more right..
@bs0409 said:If intelligent, they shld be satisfied with 1 diamond. Otherwise they are gonna die........!!!!!!!Find solutions N^4+4=p where N is natural number and p is prime number.
@bs0409 said:If intelligent, they shld be satisfied with 1 diamond. Otherwise they are gonna die........!!!!!!!Find solutions N^4+4=p where N is natural number and p is prime number.
@Subhashdec2 said:N=1 pnly??
@sails said:only 1 solution????.. not sure!!!for 2,8,4,6 power 4 last digit will be even so add 4 then it is still even..so not prime
for 3,7,9 power 4 last digit is 1 so add 4..then number will be divisible by 5..so not prime
@Estallar12 said:Method I -264 = 2^3*3*11(1,other) = 15 (total factors -1)(2,other) = 2(4,other) = 2(8,other) = 2{(2*3,11),(4*3,11),(8*3,11),(2*11,3),(4*11,3),(8*11,3),(2,3*11),(4,3*11),(8,3*11)} = 9(3,11) = 1Total = 15+6+9+1=31Method II -Condition for two divisors of any number n to be co-prime to each other(i) Let N=a^m*b^n has (m+1)(n+1)-1+mn numbers of factors which are co-prime to each other.If N=12, then as 12=2^2*3,so it has (2+1)*(1+1)-1+2=7 sets which are co-prime to eachother. They are: (1,2),(1,3),(1,4),(1,6),(1,12),(2,3),(3,4).(ii) Similarly we can deduce the formula for higher orders.If N=a^m*b^n*c^p then it will have (m+1)(n+1)(p+1)-1+mn+np+mp+3mnp co prime factors.
@bs0409 said:If intelligent, they shld be satisfied with 1 diamond. Otherwise they are gonna die........!!!!!!!Find solutions N^4+4=p where N is natural number and p is prime number.
@ravi.theja said:Find the possible number of distinct ordered pairs (a, b) of two integers a and b, such that LCM (a, b) = 264
hardik bhai even for my question also..its the same way rite??!!
Solve this:
264=a*b ,gcd(a,b)=1
How many solutions?
@bs0409 said:Solve this:264=a*b ,gcd(a,b)=1How many solutions?
@bs0409 said:Solve this:264=a*b ,gcd(a,b)=1How many solutions?
@bs0409 said:ANS=98The trick is to work backwards.Lets assume A>B>C>D>E are the thieves with A being seniormost.If only D and E are left, then D will die anyway.(Because if D dies, E gets all the 100 diamonds)Given this, if C,D,E are left, D will vote in favour of C and E will oppose C[since D does not want to goto the last round and E wants to go there]So,C will win the round anyway because of Tie.So C offers 100 for himself,0 for D,E.When B,C,D,E are left, C wants to goto next round.For D,E it doesn't matter. For B, he wants to win here only.So he offers B:98,C:0,D:1,E:1 to get votes of D and ESimilarly A will offer A:98,B:0,C:1,D:1,E:0
@ankita14 said:98 0 1 0 1
@gyrodceite said:97 is the ans..
@ankita14 said:Are you sure I had done this question before and 98 was the ans as far as I rem. logic behind 97?
@mailtoankit said:kaise bhai??