LCM based question
According to Question,
numbering of drangons :-
1. with glowing Eyes(G.E.)= 6th , 12th , 18th .......96. ( since,greatest multiple of 6 before 100 is 96).
2.Similarly, who can breathe fire(B.F.)= 4th , 8th, 12th...........100th.
3.Who shall clouds of smoke (C.S.)= 3rd, 6th ,................,99th.
4.Spikey Tail(S.T.)= 8th ,16th,................,96th.
Now,
the dragons having all four characteristics will be numbered after each interval of (24 i.e. L.C.M of 3,4,6,8) = 24th , 48th , 72nd and 96th =4 dragons (Ans.)
Now,
only two characteristics in the dragon can be:-
GE,CS = no. of dragons having these two only=numbers common to only 6 and 3 = 6,18,...........,90 =6*1 , 6*3 , .....6*15= 8 dragons
similarly,
ST,BF=8th,16th, 32nd,40th,...........80,88th.=8 dragons.
There is no other pair having only two characteristics common.
Therefore no. of dragons having two characteristics common =8+8=16 dragons
Now
No. of dragons having three characteristics common can be found similarly............:razz: