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# Ex.6.3 Q10 Squares and Square Roots Solutions - NCERT Maths Class 8

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## Question

Find the smallest square number that is divisible by each of the numbers $$8$$, $$15$$ and $$20$$

Video Solution
Squares And Square Roots
Ex 6.3 | Question 10

## Text Solution

What is known?

Divisor are $$8, 15$$ and $$20$$

What is unknown?

The smallest square that is divisible by $$8, 15$$ and $$20$$

Reasoning:

The number that will be perfectly divisible by each one of the numbers is their $$LCM$$.

Steps:

$${\rm{LCM \,of }}\,8,15,20$$

\begin{align}&2\left|\!{\underline {\,{8,15,20} \,}} \right. \\&2\left| \!{\underline {\,{4,15,10} \,}} \right. \\&2\left| \!{\underline {\,{2,15,5} \,}} \right. \\&3\left| \!{\underline {\,{1,15,5} \,}} \right. \\&5\left| \!{\underline {\,{1,5,5} \,}} \right. \\&1\left| \!{\underline {\,{1,1,1} \,}} \right. \\\end{align}

$$LCM$$ of $$8,15,20$$
\begin{align} &= 2 \times 2 \times 2\times 3 \times 5 \\&=120\end{align}

Hence prime factor $$2, 3, 5$$ do not have their respective pairs. Therefore $$120$$ is not a perfect square. If $$120$$ is multiplied by $$2 \times 3 \times 5$$ then the number obtained will be a perfect square.

Therefore, $$120 \times 2 \times 3 \times 5 = 3600$$

Thus, the perfect square number is $$3600$$ which is completely divisible by $$8, 15$$ and $$20.$$

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