Let X,Y,Z denote camel, sheep and goat
given, Px = Xa + Xb (Sum of quantities of both owners A and B)
= 2Xa (Xa = Xb)
Rx (Total revenue) = Px (Xa+Xb) = 4(Xa)^2
Now given that this revenue is just sufficient for an odd number of sheep and 1 goat, along with 2<Pz<10
4(Xa)^2 = 10Y + Pz (clearly, Xa has to be a positive integer for this to hold)
10Y can take values like 10, 30, 50, 70, 90...
i need to choose some 2<Pz<10 such that RHS is a perfect square divisible by 2
The only possible choice is Pz = 6
“Operator! Give me the number for 911!”