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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.
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