find the smallest square number which is divisible by each of 6,9 and 15
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900 is the correct answer
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we calculated LCM by prime factorisation.
LCM = 90
But, 2 and 5 are not in pair.
Therefore, to make it perfect square no. multiply 2 × 5 to LCM (90).
= 90 × 2 × 5
= 90 × 10
= 900
Hence,
Required smallest square no. that's divisible by 6, 9 and 15 = 900
LCM = 90
But, 2 and 5 are not in pair.
Therefore, to make it perfect square no. multiply 2 × 5 to LCM (90).
= 90 × 2 × 5
= 90 × 10
= 900
Hence,
Required smallest square no. that's divisible by 6, 9 and 15 = 900
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