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

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Re: If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink] New post 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] New post 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] New post 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] New post 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] New post 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] New post 22 Apr 2013, 14:54
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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 = 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] New post 25 Aug 2014, 21:21
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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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