It is currently 12 Dec 2017, 01:44

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 12 Feb 2013
Posts: 6

Kudos [?]: 19 [1], given: 18

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

### Show Tags

22 Apr 2013, 11:54
1
KUDOS
7
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

58% (01:42) correct 42% (01:32) wrong based on 171 sessions

### HideShow timer Statistics

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

Kudos [?]: 19 [1], given: 18

VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1120

Kudos [?]: 2404 [1], given: 219

Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]

### Show Tags

22 Apr 2013, 12:18
1
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.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Last edited by Zarrolou on 22 Apr 2013, 12:31, edited 1 time in total.
Edited, thanks Bunuel

Kudos [?]: 2404 [1], given: 219

Intern
Joined: 12 Feb 2013
Posts: 6

Kudos [?]: 19 [0], given: 18

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

### Show Tags

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.

Let me know if it's clear

PS:welcome to the club!

Ty nice explanation

Kudos [?]: 19 [0], given: 18

Math Expert
Joined: 02 Sep 2009
Posts: 42559

Kudos [?]: 135309 [1], given: 12686

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

### Show Tags

22 Apr 2013, 12:23
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
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.
_________________

Kudos [?]: 135309 [1], given: 12686

Math Expert
Joined: 02 Sep 2009
Posts: 42559

Kudos [?]: 135309 [1], given: 12686

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

### Show Tags

22 Apr 2013, 12:26
1
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.

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.
_________________

Kudos [?]: 135309 [1], given: 12686

Manager
Joined: 09 Apr 2013
Posts: 207

Kudos [?]: 81 [0], given: 40

Location: United States
Concentration: Finance, Economics
GMAT 1: 710 Q44 V44
GMAT 2: 740 Q48 V44
GPA: 3.1
WE: Sales (Mutual Funds and Brokerage)
Re: If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]

### Show Tags

22 Apr 2013, 12:29
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?

Kudos [?]: 81 [0], given: 40

VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1120

Kudos [?]: 2404 [0], given: 219

Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]

### Show Tags

22 Apr 2013, 12:32
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.
_________________

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

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Kudos [?]: 2404 [0], given: 219

Math Expert
Joined: 02 Sep 2009
Posts: 42559

Kudos [?]: 135309 [2], given: 12686

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

### Show Tags

22 Apr 2013, 12:35
2
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.
_________________

Kudos [?]: 135309 [2], given: 12686

Manager
Joined: 09 Apr 2013
Posts: 207

Kudos [?]: 81 [0], given: 40

Location: United States
Concentration: Finance, Economics
GMAT 1: 710 Q44 V44
GMAT 2: 740 Q48 V44
GPA: 3.1
WE: Sales (Mutual Funds and Brokerage)
Re: If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]

### Show Tags

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!

Kudos [?]: 81 [0], given: 40

Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 627

Kudos [?]: 1406 [0], given: 136

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

### Show Tags

22 Apr 2013, 15:54
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.
_________________

Kudos [?]: 1406 [0], given: 136

Non-Human User
Joined: 09 Sep 2013
Posts: 14903

Kudos [?]: 287 [0], given: 0

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

### Show Tags

25 Aug 2014, 22:21
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 287 [0], given: 0

Non-Human User
Joined: 09 Sep 2013
Posts: 14903

Kudos [?]: 287 [0], given: 0

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

### Show Tags

05 Sep 2015, 06:50
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 287 [0], given: 0

Current Student
Joined: 26 Jul 2015
Posts: 108

Kudos [?]: 24 [0], given: 66

Location: India
Concentration: Strategy, Marketing
GMAT 1: 750 Q49 V42
WE: Information Technology (Computer Software)
Re: If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]

### Show Tags

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.

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?

Kudos [?]: 24 [0], given: 66

Current Student
Joined: 20 Mar 2014
Posts: 2671

Kudos [?]: 1788 [1], given: 796

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
If n=t^3 , when n and t are positive integers and 8, 9, 10 [#permalink]

### Show Tags

05 Sep 2015, 07:54
1
KUDOS
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.

Kudos [?]: 1788 [1], given: 796

Non-Human User
Joined: 09 Sep 2013
Posts: 14903

Kudos [?]: 287 [0], given: 0

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

### Show Tags

01 Jan 2017, 22:31
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 287 [0], given: 0

Re: If n=t^3 , when n and t are positive integers and 8, 9, 10   [#permalink] 01 Jan 2017, 22:31
Display posts from previous: Sort by

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

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.