Question

find the smallest square number which is divisible by each of the numbers 6,9,15

don't give wrong answers plz ​

Answer 1

LCM (6, 9, 15) = 90

2 and 5 not occuring in pair form

But to make it a perfect square, multiply 2 and 5 to 90 (LCM)

90×2×5

900

Required smallest square number is 900

Answer 2

Step-by-step explanation:

when we find the LCM of 6,9 a and 15, we get 3,3,2,5.

multiply all these numbers along with 2 and 5.

$(3 \times 3 \times2 \times 5) \times (2 \times 5)$

now let's group them,

$3 \times 3 \times 2 \times 2 \times 5 \times 5 \\ = 3 {}^{2} \times 2 {}^{2} \times 5 {}^{2} \\ = 6 {}^{2} \times 5 {}^{2} \\ = 30 { }^{2} \\ = 900$

Thus 900 is the smallest square number divisible by 6,9 and 15

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