Question

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

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Hint: We find the LCM of three given numbers and then using prime factorization we find the prime factors. Now we have to find a number which is a square, so we try to form squares of the prime factors by multiplying the terms needed.

We know LCM of three numbers can be calculated by writing the numbers in their prime factorization.
$6 = 2 \times 3 \\ 9 = 3 \times 3 \\ 15 = 3 \times 5 \\$
So to find the LCM we will take the highest number of times whichever prime is repeated and multiply with other primes.
We have 2 only one time, 3 is multiplied two times so we take $3 \times 3$ and 5 is also there one time.
SO, LCM is $2 \times 3 \times 3 \times 5 = 90$
Now we know prime factorization of $90 = 2 \times 3 \times 3 \times 5$
To find the smallest number that is a square and is divisible by 90 we will form a number by multiplying factors to the prime factorization of 90.
So, to make $2 \times 3 \times 3 \times 5$ a square we need each number to be multiplied to itself. i.e. each prime factor should be in square.
We can write the prime factors in the form of their powers as
$\Rightarrow 2 \times 3 \times 3 \times 5 = {2^1} \times {3^2} \times {5^1}$
Now we can see 3 has a square power but 2 and 5 do not have square power. So we multiply with 2 and 5
$= {2^1} \times 2 \times {3^2} \times {5^1} \times 5 \\ = {2^2} \times {3^2} \times {5^2} \\ = 4 \times 9 \times 25 \\ = 900 \\$
So, the smallest square number that can be divided by 6, 9 and 15 is 900.

Note: Students many times make the mistake of calculating LCM wrong, as by LCM we mean least common multiple so they assume to take the least value of multiple and they take the value 3 instead of ${3^2}$which is wrong.