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By which smallest number must the following numbers be divided so that the quotient is a perfect cube?
(i) 675
(ii) 8640
(iii) 1600
(iv) 8788
(v) 7803
(vi) 107811
(vii) 35721
(viii) 243000.
To do:
We have to find the smallest number by which the given numbers must be divided so that the quotient is a perfect cube.
Solution:
(i) Prime factorisation of $675=3\times3\times3\times5\times5$
Grouping the factors in triplets of equal factors, we find that $5\times5$ is not a complete triplet.
Therefore, dividing $675$ by $5\times5=25$, we get,
$675\div25=3\times3\times3\times5\times5\div25$
$=3\times3\times3$
Hence, the smallest number by which the given number must be divided so that the quotient is a perfect cube is 25.
(ii) Prime factorisation of $8640=2\times2\times2\times2\times2\times2\times3\times3\times3\times5$
Grouping the factors in triplets of equal factors, we find that $5$ is not a complete triplet.
Therefore, dividing $8640$ by $5$, we get,
$8640\div5=2\times2\times2\times2\times2\times2\times3\times3\times3\times5\div5$
$=2\times2\times2\times2\times2\times2\times3\times3\times3$
Hence, the smallest number by which the given number must be divided so that the quotient is a perfect cube is 5.
(iii) Prime factorisation of $1600=2\times2\times2\times2\times2\times2\times5\times5$
Grouping the factors in triplets of equal factors, we find that $5\times5$ is not a complete triplet.
Therefore, dividing $1600$ by $5\times5=25$, we get,
$1600\div25=2\times2\times2\times2\times2\times2\times5\times5\div25$
$=2\times2\times2\times2\times2\times2$
Hence, the smallest number by which the given number must be divided so that the quotient is a perfect cube is 25.
(iv) Prime factorisation of $8788=2\times2\times13\times13\times13$
Grouping the factors in triplets of equal factors, we find that $2\times2$ is not a complete triplet.
Therefore, dividing $8788$ by $2\times2=4$, we get,
$8788\div4=2\times2\times13\times13\times13\div4$
$=13\times13\times13$
Hence, the smallest number by which the given number must be divided so that the quotient is a perfect cube is 4.
(v) Prime factorisation of $7803=3\times3\times3\times17\times17$
Grouping the factors in triplets of equal factors, we find that $17\times17$ is not a complete triplet.
Therefore, dividing $7803$ by $17\times17=289$, we get,
$7803\div289=3\times3\times3\times17\times17\div289$
$=3\times3\times3$
Hence, the smallest number by which the given number must be divided so that the quotient is a perfect cube is 289.
(vi) Prime factorisation of $107811=3\times3\times3\times3\times11\times11\times11$
Grouping the factors in triplets of equal factors, we find that $3$ is not a complete triplet.
Therefore, dividing $107811$ by $3$, we get,
$107811\div3=3\times3\times3\times3\times11\times11\times11\div3$
$=3\times3\times3\times11\times11\times11$
Hence, the smallest number by which the given number must be divided so that the quotient is a perfect cube is 3.
(vii) Prime factorisation of $35721=3\times3\times3\times3\times3\times3\times7\times7$
Grouping the factors in triplets of equal factors, we find that $7\times7$ is not a complete triplet.
Therefore, dividing $35721$ by $7\times7=49$, we get,
$35721\div49=3\times3\times3\times3\times3\times3\times7\times7\div49$
$=3\times3\times3\times3\times3\times3$
Hence, the smallest number by which the given number must be divided so that the quotient is a perfect cube is 49.
(viii) Prime factorisation of $243000=2\times2\times2\times3\times3\times3\times3\times3\times5\times5\times5$
Grouping the factors in triplets of equal factors, we find that $3\times3$ is not a complete triplet.
Therefore, dividing $243000$ by $3\times3=9$, we get,
$243000\div9=2\times2\times2\times3\times3\times3\times3\times3\times5\times5\times5\div9$
$=2\times2\times2\times3\times3\times3\times5\times5\times5$
Hence, the smallest number by which the given number must be divided so that the quotient is a perfect cube is 9.
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