# Find the smallest number by which the given number must be divided so that the resulting number is a perfect square.(i) 14283(ii) 1800(iii) 2904

To do :

We have to find the smallest numbers by which  each of the given numbers must be divided so that the resulting numbers are perfect squares.

Solution:

Perfect Square: A perfect square has each distinct prime factor occurring an even number of times.

(i) Prime factorisation of 14283 $=3\times3\times3\times23\times23$

$=3\times(3)^2\times(23)^2$

In order to make the pairs an even number of pairs, we have to divide 14283 by $3$, then the product will be a perfect square.

Therefore, 3 is the smallest number by which 14283 must be divided so that the resulting number is a perfect square.

(ii) Prime factorisation of 1800 $=2\times2\times2\times3\times3\times5\times5$

$=2\times(2)^2\times(3)^2\times(5)^2$

In order to make the pairs an even number of pairs, we have to divide 1800 by $2$, then the product will be a perfect square.

Therefore, 2 is the smallest number by which 1800 must be divided so that the resulting number is a perfect square.

(iii) Prime factorisation of 2904 $=2\times2\times2\times3\times11\times11$

$=2\times(2)^2\times3\times(11)^2$

In order to make the pairs an even number of pairs, we have to divide 2904 by $2\times3=6$, then the product will be a perfect square.

Therefore, 6 is the smallest number by which 2904 must be divided so that the resulting number is a perfect square.

Updated on: 10-Oct-2022

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