Question

Find the smallest square number which is divisible by each of the number $6,9$ and $15$.

Answer 1

This has to be done in two steps First find the smallest common multiple and then find the square number needed The least number divisible by each one of 6,9, and 15 is their LCM The LCM of 6,9, and 15 is $2\times 3\times 3\times 5=90$
Prime factorisation of 90 is 90 = $2\times \underset{–}{3\times 3}\times 5$
We see that prime factors 2 and 5 are not in paris Therefore 90 is not a perfect squarer
In order to get a perfet square each factor of 90 must be paired So we need to make pairs of 2 and 5 Therefore 90 should be multiplied by $2\times 5$ i.e., 10
Hence the required square number is $90\times 10$=900