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

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22 Apr 2013, 12:18

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\(8=2^3\) \(9=3^2\) \(10=5*2\)

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

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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]

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22 Apr 2013, 12:22

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.

If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]

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22 Apr 2013, 12:23

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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 \(27=3^3\), 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]

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22 Apr 2013, 12:26

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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]

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22 Apr 2013, 12:35

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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]

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22 Apr 2013, 14:42

Bunuel wrote:

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.

Ah, that makes perfect sense. Thank you for your help!

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

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25 Aug 2014, 22:21

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

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05 Sep 2015, 06:50

Hello from the GMAT Club BumpBot!

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

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05 Sep 2015, 07:23

Bunuel wrote:

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.

Hi Bunuel, Thanks for the explanation.. The rule you have used - is this a general rule? The powers of the primes of a perfect cube are always multiples of 3. For example, 1,000=23∗53 Is this true for all powers, i.e., the powers of the primes of a perfect square are always multiples of 2, for perfect (rasied to 4) they are multiples of 4 and so on?

If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]

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05 Sep 2015, 07:54

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vinod332002 wrote:

Bunuel wrote:

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.

Hi Bunuel, Thanks for the explanation.. The rule you have used - is this a general rule? The powers of the primes of a perfect cube are always multiples of 3. For example, 1,000=23∗53 Is this true for all powers, i.e., the powers of the primes of a perfect square are always multiples of 2, for perfect (rasied to 4) they are multiples of 4 and so on?

Let me try to explain.

By definition of a cube or a square , a number is a perfect square when N = \(A^{2p}\) and a number is a perfect cube when N = \(A^{3p}\), where p is an integer.

Thus for making a perfect square you need to raise the power of primes to multiples of 2 and for making a perfect cube you need to raise the power of primes to multiples of 3.

Alternately, you can see it this way: A perfect square N when taken the square root of (ie number raised to the power of 1/2), should give you an integer. In other words, if N is a perfect square, then \(\sqrt N\) = Integer. This can only happen when \(N = A^{multiple of 2}\)

Also, a perfect cube N when taken the cube root of (ie number raised to power of 1/3), should give you an integer. In other words, if N is a perfect cube, then \(\sqrt[3] N\) = Integer. This can only happen when \(N = A^{multiple of 3}\)

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