In a sports meet, the number of players in Football, Hockey and Athletics are 48,60,132, respectively.

Find the minimum number of room required, if in each room the same number of player are to be seatedand all of them being in the same sports ?

as per the requirement of the question we need the highest common factor of the numbers 48 , 60 and 132

as the number of players in each room is same and all the players ina  room belong to one sport only

prime factorisation of the numbers

48 = 2×2×2×2×3

60 = 2×2×3×5

132 = 2×2×3×11

2×2×3 is common in all the three numbers hence HCF is 2×2×3 = 12

Hence the answer is 12

you can see, the football players can be divided in to 48 = 12×4 = 4 rooms

hockey players in 60 = 12×5 = 5 rooms

athletics players in 132 = 12 × 11 = 11 rooms

hence the total number of rooms requires is 4+5+11 = 20 rooms

if we take a number higher than 12 the number of plahyers can't be equally divided