Any weighing gives 2 variants. So two weighing can give 4 variants, 3 weighing can give 8 variants and 4 weighing can give 16 variants. If the common numbers of nuggets is 4 for two heaviest nuggets exist $A_{4}^2=12$ variants. So three weighing not enought but four weighing absolutly enought. Tournir method described this gives the answer. But what about 16 nuggets? There are $A_{16}^2=16\cdot 15=240$ variants and 18 weighing which give $2^{18}$ much more numbers then needed teoreticaly. Teoreticaly should be enought 8 weighings IMHO because $2^8>240$. Is it true? Does exist better method for solving this problem (less then 18 weighings)?
Remark. From the problem we should find two heaviest nuggets (without it's sorting) it's mean that exist not 240 but 120 different variants for 2 heaviest nuggets.