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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 \(20.\)