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

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

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22 Apr 2013, 10:54
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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
[Reveal] Spoiler: OA
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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, 11:18
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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

PS:welcome to the club!
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Last edited by Zarrolou on 22 Apr 2013, 11:31, edited 1 time in total.
Edited, thanks Bunuel
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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, 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.

Let me know if it's clear

PS:welcome to the club!

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

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22 Apr 2013, 11:23
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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.

P.S. Please post PS questions in PS forum.
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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, 11:26
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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.

Let me know if it's clear

PS:welcome to the club!

The least value of n is (2*3*5)^3.

Also, 225=3^2*5^2 not 5^3.
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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, 11:29
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?
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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, 11:32
Bunuel wrote:
The least value of n is (2*3*5)^3.

Also, 225=3^2*5^2 not 5^3.

Of course, you're right.
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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, 11:35
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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.
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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, 13:42
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!
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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, 14:54
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 = 8p,n = 9q and n = 10r.Thus,$$n^3 = t^6 = 2^4*3^2*5*k$$[k is some positive integers]

Thus, for t to be a positive integer, k must atleast have the value = $$2^2*3^4*5^5$$

Thus,$$n = t^3 = 2^3*3^2*5^2*y$$ [y is a positive integer]

225 is definately a factor of n.

C.
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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, 05:50
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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, 06:23
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.

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

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05 Sep 2015, 06:54
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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.

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 $$N = A^{multiple of 3}$$

Hope this helps.
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If n=t^3 , when n and t are positive integers and 8, 9, 10   [#permalink] 05 Sep 2015, 06:54
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