Question

find the smallest square number that must be divisible by each of the number 6, 9 and 15​

Answer 1

$\huge \bf \mathbb \ {Answer}$

$\bf \: smaller$ $\bf \: square$$\bf \: number\bf \: divisible$ $\bf \: by$ $\bf 6$$,$$\bf9$$,$$\bf \:1 5$$\bf \: =$$\bf \: L.C.M$$\bf \: of$ $\bf \: 6$$,$$\bf \: 9$$,$$\bf \: 15$. $\qquad$ $\bf or$$\bf \: multiple$$\bf \: of$$\bf \: L.C.M$

$\bf \: Finding \: L.C.M \: of \: 6, 9,1 5$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{array}{r | l}2 &6,9,15\: \: \\3&3,9 ,15\: \: \\ 3&1,3,5 \: \: \\ 5& \: 1,1,5\: \: \\&1 ,1,1 \: \: \: \\\: \: \: \: \ \: \end{array}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\bf L.C.M \:of 6,9,15 = 2×3×3×5$

$\bf \: L.C.M \: of \: 6,9,15 = 2,3,3,5$

$\: \: \: \: \: \: \qquad \: \qquad \qquad \: \bf \: = 90$

$\bf \: Now \: checking \: if \: 90 \: is \: a \: perfect \: sqaure \: or \: not$

$\bf \: Checking \: if \: 90 \: is \: a \: \: perfect \: square \: number. \: </p><p>$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{array}{r | l}2 &90\: \: \\3 &45\: \: \\ 3&15 \: \: \\ 5& \: 5\: \: \\&1 \: \: \: \\\: \: \: \: \ \: \end{array}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\bf \: 90 \: = \: 2 \times \underline{ 3 \times 3} \times 5$

$\bf \: Since \: 2 \: and \: \: 5 \: do \: not \: occur \: in \: pairs$

$\bf \: \therefore \: 90 \: is \: not \: a \: perfect \: square \:$

$\bf \: so , \: We \: Multiply \: by \: 2 \: and \: 5 \: to \: make \: pairs </p><p>$

$\bf \: \: so , \: our \: multiply \: Becomes </p><p>$

$\bf \: 90 \times 2 \times 5 = 2 \times 3 \times 3 \times 5 \times 2 \times 5$

$\bf \: 900 \: = \: \underline{2 \times 2} \times \underline{ 3 \times 3} \times \underline{ 5 \times 5}$

$\bf \: Now , \: it \: becomes \: a \: perfect \: sqaure \:$

$\bf \: Thus \: required \: number \: is \: 900$

Answer 2

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