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Re: If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]
22 Apr 2013, 11:18

1

This post received KUDOS

8=2^3 9=3^2 10=5*2

So n is formed by 2,3,5. Because both t and n are positive integers we can write n=(2*3*5)^3

Now to the question: which of the following must be a factor of n? All the answers (except the correct one) are not possible combinations of (2*3*5)^3 so they are not factors Only 225=3^2*5^2 is in n, so C is the answer.

Let me know if it's clear

PS:welcome to the club! _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]
22 Apr 2013, 11:22

Zarrolou wrote:

8=2^3 9=3^2 10=5*2

So n is formed by 2,3,5. Because both t and n are positive integers we can write n=(2*3*5)^3

Now to the question: which of the following must be a factor of n? All the answers (except the correct one) are not possible combinations of (2*3*5)^3 so they are not factors Only 225=5^3 is in n, so C is the answer.

Re: If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]
22 Apr 2013, 11:23

Expert's post

alex90 wrote:

If n=t^3 , when n and t are positive integers and 8, 9, 10 are each factors of n, which of the following must be a factor of n?

A. 16 B. 175 C. 225 D. 275 E. 625

n=t^3=integer^3 means that n is a perfect cube. A perfect cube is an integer that is the cube of an integer. For example 27=3^2, is a perfect cube.

The powers of the primes of a perfect cube are always multiples of 3. For example, 1,000=2^3*5^3: the powers of 2 and 5 are multiples of 3.

Now, since 8=2^3, 9=3^2, and 10=2*5 are factors of n, then 2, 3, and 5 are factors of n, therefore 2^3=8, 3^3=27 and 5^3=125 must be the factors of n.

A. 16=2^4: not necessarily a factor of n. B. 175=5^2*7: not necessarily a factor of n. C. 225=3^2*5^2: since 3^3=27 and 5^3=125 are the factors of n, then 225 is also a factor of n. D. 275=5^2*11: not necessarily a factor of n. E. 625=5^4: not necessarily a factor of n.

Answer: C.

P.S. Please post PS questions in PS forum. _________________

Re: If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]
22 Apr 2013, 11:26

1

This post received KUDOS

Expert's post

Zarrolou wrote:

8=2^3 9=3^2 10=5*2

So n is formed by 2,3,5. Because both t and n are positive integers we can write n=(2*3*5)^3

Now to the question: which of the following must be a factor of n? All the answers (except the correct one) are not possible combinations of (2*3*5)^3 so they are not factors Only 225=5^3 is in n, so C is the answer.

Re: If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]
22 Apr 2013, 11:35

1

This post received KUDOS

Expert's post

dave785 wrote:

This one doesn't quite make sense to me.

I factored it down to 2*2*2 (8), 3*3 (9) and 2*5 (10).

Those are factors of N, not T.

The 2 in (10) i cancelled out since we're already using one in (8), and then multiplied it out and got 360, which wasn't an answer choice.

Is there a mathematical rule that limits the possible factors of a perfect cube?

The powers of the primes of a perfect cube are always multiples of 3. For example, 1,000=2^3*5^3: the powers of 2 and 5 are multiples of 3.

Now, since 8=2^3, 9=3^2, and 10=2*5 are factors of n, then 2, 3, and 5 are factors of n, therefore 2^3=8, 3^3=27 and 5^3=125 must be the factors of n. _________________

Re: If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]
25 Aug 2014, 21:21

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