Question

what is the smallest number by
which 8640 must be divided
so that
the quotient is a perfect
cube ?
​

Answer 1

Question

what is the smallest number by  which 8640 must be divided  so that  the quotient is a perfect  cube?

$\rule{300}{1}$

Answer

First we need to prime factorize 8604.

$\begin{array}{r | l} 2 & 8640 \\ \cline{2-2} 2 & 4320 \\ \cline{2-2} 2 & 2160 \\ \cline{2-2} 2 & 1080\\ \cline{2-2} 2& 540 \\ \cline{2-2} 2 & 270\\\cline{2-2} 3 &135 \\ \cline{2-2} 3 & 45\\\cline{2-2} 3 & 15 \\ \cline{2-2} & 5\\ \end{array}$

∴ 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5

Now we will have to group the product of primes with 3 in each group.

(2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × 5

∴8640 is not a perfect cube because '5' doesn't lie in any group.

∴ We must divide 5 from 8640 to obtain a perfect cube.

Let's verify if we get a perfect square after dividing 5 from 8640.

8640 ÷ 5 = 1728

Now we must do prime factorization for 1728.

$\begin{array}{r | l} 2 & 1728 \\ \cline{2-2} 2 & 864 \\ \cline{2-2} 2 & 432 \\ \cline{2-2} 2 & 216 \\ \cline{2-2} 2 & 108 \\ \cline{2-2} 2 & 54 \\\cline{2-2} 3 & 27 \\ \cline{2-2} 3 & 9\\\cline{2-2} 3 &3 \\ \cline{2-2} &1 \\ \end{array}$

∴ 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

Now we will have to group the product of primes with 3 in each group.

(2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3)

∴1728 is a perfect cube because all numbers lie in any groups.

∴ We must divide 5 from 8640 to obtain a perfect cube.

$\rule{300}{1}$

Answer 2

Answer:

5

Step-by-step explanation:

2+1+1+1