find the each case the smallest perfect square divisible by the given number
(1) 6,8,10 and 12.
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Take LCM of (6,8,12 and 15) = 120
Resolving 120 into prime factors, we get
120 = 2*2*2*3*5
Here 2 is grouped in pairs of equal factors. But 2, 3 and 5 are not grouping in pairs of equal factors.
Let us multiply 2, 3 and 5 , we get a grouped in pairs of equal factors.
120*2*3*5 = 2*2*2*2*3*3*5*5
3600 = 2*2*2*2*3*3*5*5
Now 3600 is perfect square that is divisible by 6, 8, 12, and 15.
HOPE IT HELP YOU …ASK EVERY QUESTION WHICH YOU HAVE DOUBT ❤️
Resolving 120 into prime factors, we get
120 = 2*2*2*3*5
Here 2 is grouped in pairs of equal factors. But 2, 3 and 5 are not grouping in pairs of equal factors.
Let us multiply 2, 3 and 5 , we get a grouped in pairs of equal factors.
120*2*3*5 = 2*2*2*2*3*3*5*5
3600 = 2*2*2*2*3*3*5*5
Now 3600 is perfect square that is divisible by 6, 8, 12, and 15.
HOPE IT HELP YOU …ASK EVERY QUESTION WHICH YOU HAVE DOUBT ❤️
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