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

We will first find out the L.C.M of 6, 9 and 15.

L.C.M of 6, 9, 15 = 2 × 3 × 3 × 5 = 2 × 32  × 5 = 90

Since, we need to find the smallest square number divisible by 6, 9 and 15 and in above prime factorization of the numbers we observe that 2 and 5 does not appear in pair, therefore we multiply the L.C.M, 90 by 2 × 5.

Hence,

The required smallest square number = 90 × 2 × 5 = 900.

  • 34

smallest square number = l.c.m of 6,9,15 = 90

  • -3
 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 × 3 × 3 × 5 = 90.
Prime factorisation of 90 is 90 = 2 × 3 × 3 × 5.
We see that prime factors 2 and 5 are not in pairs. Therefore 90 is not a perfect
square.
In order to get a perfect 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 × 5, i.e., 10.
Hence, the required square number is 90 × 10 = 900.
  • 15

LCM =  

2 | 6,9,15
3 | 3,9,15
3 | 1,3,5
5 | 1,1,5
   | 1,1,1

so, 2x3x3x5 = 90
hence, 90 is smallest square number divisible by 6,9,15 

  • -8

Thanks,all of you...!!!

  • 1
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