What is the smallest natural number by which 135 must be divided so that the quotient is a perfect cube?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5
In a perfect cube every prime factor appears in groups of three. Dividing by the factors left over outside such groups leaves a perfect cube.
\(135=3\times3\times3\times5\)
The three 3s form one group; the 5 is left over
\(135\div5=27=3^3\), a perfect cube
Why the other options are incorrect
Option (1) 3 — Incorrect. \(135\div3=45=3\times3\times5\), which has no complete group of three.
Option (3) 7 — Incorrect. 7 is not a factor of 135, so the quotient is not even a whole number.
Option (4) 9 — Incorrect. \(135\div9=15=3\times5\), which is not a perfect cube.