[RESOLVED] Perfect cube???

[Yunie]

Expressed as the product of prime factors


540 = 2^2 x 3^2 x 5 and 72 = 2^3 x 3^2


Use these results to find

the smallest integer, t, such that 540t is a perfect cube.


I have looked through the working solutions but I don't understand the steps on why it is been done in that way. Can anyone explain it to me or maybe you could do it in another way so that I can understand it better? Thanks a lot.


Working solutions:

540t = 2^2 x 3^2 x 5 x (2x5^2)
=> t = 2 x 5^2
t = 50


Please answer this thread asap. Thanks a lot. :)

[jemidiah]

I think there's a typo in your question, since 540 = 22 33 5

Anywho, there are two basic facts you need to know to solve this problem:

1. First, prime factorizations are unique to all positive integers greater than 1. That is, there is only one list of all the prime factors of a given positive integer (note I said "list"; the order isn't important). The prime factorization of a number is a list of all the primes (an infinite number of numbers) raised to some integer powers. For example, 858 = 2*3*11*13, and 11154 = 858*13 = 2*3*11*132. 1 isn't considered prime (otherwise we'd have to make exceptions for it all the time), so the list is unique. Also note that all other primes in existence are taken to the 0th power--and any number to the 0th is 1, so they collapse out of the multiplication and aren't normally written. That is, 11154 = 2*3*50*70*11*132*170*...

2. A perfect cube is a number C such that C = c3 for some integer c (I've included a less formal definition later).


Now, to answer the question.
You can think of multiplying two different numbers as multiplying their prime factorizations. So, 9*8 is also 3*3 * 2*2*2 or 32 * 23. In the same way, 540*t can be thought of as multiplying prime factors.

We know the prime factorization of 540, and in math you always try to plug in as much (useful) information as you can to get a solution, so let's plug that in:
540t = 22 33 5 * t. That's as far as we can go without another idea--let's get one.

We need 540t to be a perfect cube--but what does that mean? What is a perfect cube? Well, it's an integer that can be cube-rooted without returning a fraction. Or, it's an integer (call the perfect cube C) that has a corresponding integer (call this number c) such that c*c*c = C, or C = c3. Note that I've used big C and little c so that c <= C [since cube rooting an integer always results in a smaller (or equal, in the case of cube rooting 1) integer]. I like using visual aids like this--they help me remember what a particular variable/constant is for.

So? Does this help us? Yes, but we have to plug in the fact that all positive integers have a unique prime factorization--that is, any integer x > 1 can be written as 2a 3b 5c ... for some set of numbers a, b, c, ... (note that any of these numbers can be 0, and any prime number to the 0th power is 1, which disappears from the prime factorization list when it's normally written) (also note that I'm using the elipses [the ...] to mean, "and so on"). So, the number c--that is, the cube root of C, where C is our perfect cube--can be written as a list of primes taken to various powers. Algebraically...

c = 2c2 3c3 5c5 ... for some set of constants c2, c3, c5, ...

Alrighty! So, we know c = 2c2 3c3 5c5 ... and c3 = C. Let's combine these and hope something helps us solve our earlier problem.

Substituting the first equation in for c in the second equation, you get (with some algebra)
C = (2c2 3c3 5c5 ...)3 = 23*c2 33*c3 53*c5 ...

Note: I used the fact that (a^b * c^d)^e = (a^(b*e)) * (c^(d*e)) here. Example: (2^3 * 5)^2 = 1600 = 2^6 * 5^2

What does this mean? It means that given ANY perfect cube, we [i]know that its prime factorization uses powers that are only factors of 3 (0, 3, 6, 9, 12, ...), since prime factorizations are unique! So our perfect cube 540t must also have a prime factorization that uses factors of 3 as powers!

Earlier, we found that 540t = 22 33 5(1) * t. What prime factorization for t would make the prime factors in 540t all raised to powers that are multiples of 3? Remember the identity: a^b * a^c = a^(b+c). Example: 3^4 * 3^2 = 729 = 3^6.

Hopefully you've seen it by now: make t = 21 30 52 so that 540t = 22+1 33+0 51+2 = 23 33 53. Note that you could also use t= 24 33 55 or any combination of powers that make the exponents of the prime factors of 540t multiplies of 3--but, the first one I listed is the smallest such number.

Thus, t = 2*(1)*5*5 = 2*25 = 50. Done!


P.S. Also, perhaps the reason you were confused when given the "working solution" you posted above was that the solution doesn't actually prove anything--it just tells you the answer. So, it probably doesn't have a good chance of explaining the answer to you :P

[obi1kenobi]

I think there's a typo in your question, since 540 = 22 33 5


I agree, there must be a typo or the solution is incorrect.

We need 540t to be a perfect cube--but what does that mean? What is a perfect cube? Well, it's an integer that can be cube-rooted without returning a fraction. Or, it's an integer (call the perfect cube C) that has a corresponding integer (call this number c) such that c*c*c = C, or C = c3. Note that I've used big C and little c so that c <= C [since cube rooting an integer always results in a smaller (or equal, in the case of cube rooting 1) integer].

A non-perfect cube will NOT return a fraction, it will return a irrational number, which by definition is an infinitely long decimal number which CANNOT be written as a fraction.

irrational number: real number that is not rational: any real number that cannot be expressed as the exact ratio of two integers, e.g. √2 and π (pi)

Microsoft® Encarta® 2007. © 1993-2006 Microsoft Corporation. All rights reserved.

[Yunie]

Hey thanks to both jemidiah and obi1kenobi for replying to my thread. Thanks a lot! I really do appreciate both of you guys' help! :)

To jemidiah, I have 1 more question to ask you. You said that given ANY perfect cube, its prime factorization uses power that are only factors of 3. So, if the question asks for perfect square (a specific number to the power of 2), the prime factorization must use only power that are factors of 2, right? Thanks again! :)

[obi1kenobi]

Hey thanks to both jemidiah and obi1kenobi for replying to my thread. Thanks a lot! I really do appreciate both of you guys' help! :)

To jemidiah, I have 1 more question to ask you. You said that given ANY perfect cube, its prime factorization uses power that are only factors of 3. So, if the question asks for perfect square (a specific number to the power of 2), the prime factorization must use only power that are factors of 2, right? Thanks again! :)

Why not to me, am I not qualified to answer it? :D I won a gold medal on the Junior Balkan Mathematical Olympiad this year, you know :cool: (and it's not a joke). You are right, if the question asks for a perfect square, the prime factorization must contain only powers that are factors of 2. The same applies to any integer n: if n-th root of a number is an integer, then the number's prime factorization uses only powers that are factors of n, and vice versa. If you have any more questions, feel free to ask. ;)

[Yunie]

Why not to me, am I not qualified to answer it? :D I won a gold medal on the Junior Balkan Mathematical Olympiad this year, you know :cool: (and it's not a joke). You are right, if the question asks for a perfect square, the prime factorization must contain only powers that are factors of 2. The same applies to any integer n: if n-th root of a number is an integer, then the number's prime factorization uses only powers that are factors of n, and vice versa. If you have any more questions, feel free to ask. ;)


Haha. Of course you are qualified to answer my question. Hahaha. :D Wow, a gold medal on the Junior Balkan Mathematical Olympiad! That's great and cool! For now, I don't think I have any questions to ask in regards to this question...Maybe you could help me solve my other questions in the other threads? Thanks a lot for answering my questions and clearing my doubts. Thanks a lot! I really appreciate it! :thumb:

[obi1kenobi]

You're welcome, any time. If you like my explanations, please rate my posts. Thanks.

[Yunie]

Thanks. I really like your explanations and I have already rated your posts. Cheers! :thumb: