Cubes and Cube Roots Test - 19

Prime factorising $$1440$$, we get,

$$ 1440 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5$$

          $$= 2 ^5 \times 3 ^2 \times 5^1$$.

We know, a perfect cube has multiples of $$3$$ as powers of prime factors.

Here, number of $$2$$'s is $$5$$, number of $$3$$'s is $$2$$ and number of $$5$$'s is $$1$$.

So we need to multiply another $$2$$, $$3$$ and $$5^2$$ in the factorization to make $$1440$$ a perfect cube.

Hence, the smallest number by which $$1440$$ must be multiplied to obtain a perfect cube is $$2\times 3 \times 5^2=150$$.

$$\therefore$$ The sum of digits of the smallest number is $$= 1+5+0 = 6$$.

Hence, option $$B$$ is correct.

Prime factorising $$9000$$, we get,

$$ 9000 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 5$$

          $$= 2^3 \times 3 ^2 \times 5^3$$

We know, a perfect cube has prime factors in powers of $$3$$.

Here, number of $$2$$'s is $$3$$, number of $$3$$'s is $$2$$ and number of $$5$$'s is $$3$$.

So we need to remove $$3^2$$ from the factorization to make $$9000$$ a perfect cube.

Hence, the smallest number by which $$9000$$ must be divided to obtain a perfect cube is $$3^2=9$$.

Prime factorising $$25$$, we get,

$$25=5\times 5=5^2$$.

We know, a perfect cube has multiples of $$3$$ as powers of prime factors.

Here, number of $$5$$'s is $$2$$.

So we need to multiply another $$5$$ in the factorization to make $$25$$ a perfect cube.

Hence, the smallest number by which $$25$$ must be multiplied to obtain a perfect cube is $$5$$.

Prime factorising $$1296$$, we get,

$$1296= 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$$

          $$= 2^4 \times 3 ^4$$.

We know, a perfect cube has multiples of $$3$$ as powers of prime factors.

Here, number of $$2$$'s is $$4$$ and number of $$3$$'s is $$4$$.

So we need to divide $$2$$ and $$3$$ from the factorization to make $$1296$$ a perfect cube.

Hence, the smallest number by which $$1296$$ must be divided to obtain a perfect cube is $$2 \times 3=6$$.

Prime factorising $$4096$$, we get,

$$4096 = 2\times2\times2\times2\times2\times2\times$$ $$2\times2\times2\times2\times2\times2$$ $$ = 2^{12}$$.

We know, a perfect cube has multiples of $$3$$ as powers of prime factors.

Here, number of $$2$$'s is $$12$$, which is a multiple of $$3$$.

Therefore, $$4096$$ is a perfect cube.