Question

which is the smallest number by which 8112 must be divided to get a perfect square?​

Answer 1

We have to find the smallest number by which 8112 should be divided to get a perfect square. Let's prime factorize 8112 to find a number which does not exist in pair:

$\Large{ \begin{array}{c|c} \tt 2 & \sf{ 8112} \\ \cline{1-2} \tt 2 & \sf { 4056} \\ \cline{1-2} \tt 2 & \sf{ 2028} \\ \cline{1-2} \tt 2 & \sf{ 1014} \\ \cline{1-2} \tt 3 & \sf{ 507 }\\ \cline{1-2} \tt 13 & \sf{ 169 }\\ \cline{1-2} \tt 13 & \sf{ 13 }\\ \cline{1-2} & \sf{ 1} \end{array}}$

8112 = 2 x 2 x 2 x 2 x 3 x 13 x 13

The factor 3 does not occur in pair.

Thus, '3' is the smallest number by which 8112 should be divided to get a perfect square.

• 8112/3 = 2 x 2 x 2 x 2 x 13 x 13
• 2704 = (2 x 2 x 13)²
• 2704 = 52²

2704 is a perfect cube.

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