By which smallest number must the following number be divided so that the quotient is a perfect cube? 7803 On factorising 7803 into prime factors, we get: \[7803 = 3 \times 3 \times 3 \times 17 \times 17\] On grouping the factors in triples of equal factors, we get: \[7803 = \left\{ 3 \times 3 \times 3 \right\} \times 17 \times 17\] It is evident that the prime factors of 7803 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 7803 is a not perfect cube. However, if the number is divided by (\[17 \times 17 = 289\]), the factors can be grouped into triples of equal factors such that no factor is left over. Thus, 7803 should be divided by 289 to make it a perfect cube. Concept: Concept of Cube Root Is there an error in this question or solution? |