Question

How many perfect cubes divide evenly into 13,824?

Please explain.

Answer 1

First step would be to express 13824 and the product of primes - can you do that?

Answer 2

So, you're supposed to multiply prime numbers to get 13824?

Answer 3

yes, e.g.:\[18=2\times3^2\]

Answer 4

any number can be expressed as the product of powers of prime numbers

Answer 5

its called the prime factorisation of a number

Answer 6

What are you supposed to do after that?

Answer 7

lets take a smaller example - 216. This can be written as:\[216=2^3\times3^3\]Can you see now how to get all perfect cubes that divide evenly into 216?

Answer 8

Yes.

Answer 9

So try the same thing with the number you are given

Answer 10

Okay, but one of the things confusing me is that the answer is supposed to be 7.

Answer 11

what prime factorisation did you get for 13824?

Answer 12

I got 2\[2^{9} and 3^{3}\]

Answer 13

correct, so:\[13824=2^9\times3^3\]Now you need to work out how many cubes can be formed from this - what do you get?

Answer 14

What do you mean by how many cubes?

Answer 15

e.g. \(3^3\) is a perfect cube that would divide evenly into 13824

Answer 16

Oh, so for 2^9, would you break that into 2 further cubes that would be 2^3 and 2^3?

Answer 17

well from \(2^9\) you can see that you have these perfect cubes: \(2^3\), \((2^2)^3=4^3\)

Answer 18

similarly, you can use combinations of 2 and 3 to get other cubes

Answer 19

and of course \(2^9\) itself is also a perfect cube

Answer 20

So, for this question then, you have 2^3, 2^3 and 3^3, do you just add the numbers then like, 2,2, and 3?

Answer 21

no

Answer 22

\(2^3\) is one cube that divides evenly into 13824
\((2^2)^3=4^3\) is another cube that divides evenly into 13824
\((2^3)^3=8^3\) is another cube that divides evenly into 13824
\(3^3\) is another cube that divides evenly into 13824
now find the others that can be formed by combining the two prime factors 2 and 3 to get perfect cubes

Answer 23

I'm not sure...

Answer 24

HINT: One of them would be \((2\times3)^3=6^3\)

Answer 25

I'm sorry, but I still don't understand how to to solve it. I get what you're saying with prime factorization and stuff, but I don't get how to find the answer.

Answer 26

np - I must be explaining it badly - let me try another approach...

Answer 27

we know that:\[13824=2^9\times3^3\]and we are asked to find all perfect cubes that divide evenly into this number.

So, we are supposed to find cubes, \(x^3\), such that:\[13824=x^3\times\text{some other number}\]does that make sense so far?

Answer 28

Yupp.

Answer 29

good - so a simple one would be:\[13824=2^3\times(2^6\times3^3)\]giving us \(x^3=2^3\)

make sense?

Answer 30

Right.

Answer 31

so basically we are being asked to work out in how many different ways can we write the product \(2^9\times3^3\) such that it is the product of a cube and some other number

Answer 32

the examples we have found so far are: