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 \(20\)

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 \(20.\)