Question

Find the smallest number by which 185220 should be divided to make it a perfect cube

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,

The given number or value , is ; "185220"

By the method of prime facotsisarion, that is , prime factors or prime numbers of this particular value "185220" ,,,

$= > 185220 = > 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 7 \times 7 \times 7$

Now , the groups of three numbers , that is, into a group of three , are , "3" , "7" , "2" and "5" are not in the group unlike the other two factors (present in the group) , so , in order to obtain the smallest least number or smallest number which should get divided to make a perfect cube will be the imperfect cubic values itself , that is ,

$= > "2" \: \: "2" \: \: and \: \: "5" \: \: \: are \: the \: ones \: to \: be \: taken \: into \: special \: \: consideration.$

Firstly let's see the importance of a perfect cube whole numbers into groups of three as a divisor,,,,

$= > {(3)}^{3} \times {(7)}^{3} \\ \\ = > {(3 \times 7)}^{3} \\ \\ Since \: , \: \: it \: \: is \: \: a \: whole \: \: number \: \: it \: can \: \: get \: a \: \: cube \: \: root \: for \: \: simplification \\ \\ = > \sqrt[3]{ {(3 \times 7)}^{3} } \\ \\ = > 3 \times 7 \\ \\ = > 21$

Now the division of the two gives us a imperfect cube , it is not fitting the criteria, hence the special consideration here , will be , the imperfect cubes , and , since they are already imperfect and don't produce a whole value they are going to be taken already standard forms , or , default prime factors, that is ,,,

$= > 2 \times 2 \times 5 \: \: Are \: \: the \: \: prime \: \: factors \: \: not \: \: occuring \\ in \: \: groups \: \: of \: \: three \\ \\ = > 2 \times 2 \times 5 \\ \\ = > 20$

Dividing to check the whole number divisibility of the number or value "20" with the main number of "185220" ,,, that is ,,,,

$= > \frac{185220}{20} \\ \\ = > \frac{18522}{2} \\ \\ = > 9261$

Now checking the cube root of obtained number or value "9261" in order to confirm it's the right answer ,,,,

$= > \sqrt[3]{9261} \\ \\ = > 21$

Therefore the number or least value of $\textbf{"20"}$ is the one making the value or number "185220" appear as a perfect cube after getting it divisible by it

HOPE IT HELPS YOU AND CLEARS YOUR DOUBTS!!!!!!

Answer 2

Find the prime factors of 185220:

185220 = 2² x 3³ x 5 x 7³

Define k:

Let k be the number that is needed t0 divide the number to be a perfect cube

Solve k:

Perfect cube ⇒ the power has to be in multiple of 3s.

185220 = 2² x 3³ x 5 x 7³

185220/k = (2² x 3³ x 5 x 7³) / ( 2² x 5)

k = 2² x 5

k = 20

Answer: The number is 20