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Answer :
The smallest number divisible by 8, 15 and 20 is equal to their LCM.
\(\qquad \begin{array}{c|lcr}
2 & 8 \;,\;15\;,\;20\\
\hline
2 & 4 \;,\;15\;,\;10\\
\hline
2 & 2 \;,\;15\;,\;5\\
\hline
3 & 1 \;,\;15\;,\;5\\
\hline
5 & 1 \;,\;5\;,\;5\\
\hline
& 1\;,\;1\;,\;1\\
\end{array}
\)
LCM =\( 2 \times 2 \times 2 \times 3 \times 5 = 120\)
Here, 2, 3 and 5 have no pair.So they are multiplied to get the least number divisible by 8,15 and 20.
The required smallest square number =\( 120 \times 2 \times 3 \times 5 = 120 \times 30 = 3600\)