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Question
By which smallest number must the following number be divided so that the quotient is a perfect cube?
243000
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Solution
On factorising 243000 into prime factors, we get:
\[243000 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5\]
On grouping the factors in triples of equal factors, we get:
\[243000 = \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 3 \times 3 \times 3 \right\} \times 3 \times 3 \times \left\{ 5 \times 5 \times 5 \right\}\]
It is evident that the prime factors of 243000 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243000 is a not perfect cube. However, if the number is divided by (\[3 \times 3 = 9\]), the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, 243000 should be divided by 9 to make it a perfect cube.
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