It is currently 18 Mar 2018, 19:54

### 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 (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit

Author Message
TAGS:

### Hide Tags

Manager
Joined: 01 Sep 2012
Posts: 125
If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

25 Feb 2013, 14:13
3
KUDOS
18
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

54% (00:58) correct 46% (01:18) wrong based on 550 sessions

### HideShow timer Statistics

If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both positive integers, how many different possible values of n are there?

A. 1
B. 2
C. 3
D. 4
E. 6

[Reveal] Spoiler:
So what I did is basically making 35^n = (5^n)(7^n)

Thought the answer is B but i was wrong.
Can anyone explain?
[Reveal] Spoiler: OA

_________________

If my answer helped, dont forget KUDOS!

IMPOSSIBLE IS NOTHING

Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4679
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

25 Feb 2013, 15:31
3
KUDOS
Expert's post
1
This post was
BOOKMARKED
roygush wrote:
If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both positive integers, how many different possible values of n are there?

A.1
B.2
C.3
D.4
E.6

So what I did is basically making 35^n = (5^n)(7^n)

Thought the answer is B but i was wrong.
Can anyone explain?

Dear Roygush

I must say, starting from 35^n = (5^n)(7^n), it's not clear to me how you wound up with only two possibilities. Your reasoning is not clear to me. I would say --- start on the left side ---

The factor of (3^4) doesn't play into the (35^n) at all --- that will have to go into the x, no choice.

We have three factors of 7, so that means we could have as many as three powers of 35 ------ we could have 35^1, 35^2, or 35^3. We can't have 35^0 = 1, because even though 1 is positive integer, that would make n = 0, and zero is not a positive integer. So, those three values of n, 1 and 2 and 3, are the only possibilities. Therefore, there are three possibilities.

n = 1 -------> x = (3^4)(5^5)(7^2)
n = 2 -------> x = (3^4)(5^4)(7^1)
n = 3 -------> x = (3^4)(5^3)

Does all that make sense?

Mike
_________________

Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Intern
Joined: 08 Dec 2012
Posts: 7
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

25 Feb 2013, 18:45
2
KUDOS
(3^4) (5^3) (5^3) (7^3) = (3^4) (5^3) (35^3) => n = 3, x = (3^4) (5^3)
Math Expert
Joined: 02 Sep 2009
Posts: 44298
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

26 Feb 2013, 03:02
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both positive integers, how many different possible values of n are there?

A. 1
B. 2
C. 3
D. 4
E. 6

Notice that the power of 7 in LHS is limiting the value of n, thus n cannot be more than 3 and since n is a positive integer, then n could be 1, 2, or 3.

_________________
Manager
Joined: 01 Sep 2012
Posts: 125
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

26 Feb 2013, 08:40
thank you all.
Its clear now and i should have thought about it myself.
_________________

If my answer helped, dont forget KUDOS!

IMPOSSIBLE IS NOTHING

Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4679
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

26 Feb 2013, 11:33
This question turned out to be relatively easy, but it inspired me to create a similar question that's a little more challenging:
if-a-and-b-are-positive-integers-and-147953.html

Mike
_________________

Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

Manager
Joined: 04 Mar 2013
Posts: 84
Location: India
Concentration: General Management, Marketing
GPA: 3.49
WE: Web Development (Computer Software)
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

06 Jul 2013, 12:34
Bunuel wrote:
If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both positive integers, how many different possible values of n are there?

A. 1
B. 2
C. 3
D. 4
E. 6

Notice that the power of 7 in LHS is limiting the value of n, thus n cannot be more than 3 and since n is a positive integer, then n could be 1, 2, or 3.

hi, i differ on this,

here is the reason

35^0 is also possible as we are not given any limitation to x

we can have values of 35^0, 1, 2, 3, so n can take 4 values .......... and the answer would be 4 D
Math Expert
Joined: 02 Sep 2009
Posts: 44298
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

06 Jul 2013, 12:37
krrish wrote:
Bunuel wrote:
If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both positive integers, how many different possible values of n are there?

A. 1
B. 2
C. 3
D. 4
E. 6

Notice that the power of 7 in LHS is limiting the value of n, thus n cannot be more than 3 and since n is a positive integer, then n could be 1, 2, or 3.

hi, i differ on this,

here is the reason

35^0 is also possible as we are not given any limitation to x

we can have values of 35^0, 1, 2, 3, so n can take 4 values .......... and the answer would be 4 D

Please read the stem carefully: "x and n are both positive integers..." 0 is NOT a positive integer.
_________________
SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1838
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

26 Feb 2014, 23:20
1
KUDOS
Re-arrange the equation:

x = 3^4 . 5^6. 7^3 / 35^n

n may be 0,1,2 3........ however n is +ve, so can be 1,2,3

_________________

Kindly press "+1 Kudos" to appreciate

Senior Manager
Joined: 20 Dec 2013
Posts: 266
Location: India
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

08 Mar 2014, 10:32
Option C.
RHS:7^n * 5^n * x
On the LHS,7 has max. Power of 3
Since n is a +ve integer,it can't be zero.So n=1,2,&3 only.

Posted from my mobile device
Intern
Joined: 22 Feb 2014
Posts: 2
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

27 Mar 2014, 15:37
How is this problem solved? I don't understand why the powers of 7 limits the answer choices. Maybe I am thinking about this the wrong way.
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4679
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

27 Mar 2014, 16:14
1
KUDOS
Expert's post
frenchwr wrote:
How is this problem solved? I don't understand why the powers of 7 limits the answer choices. Maybe I am thinking about this the wrong way.

Dear frenchwr,
I'm happy to respond.

How well do you understand the idea of the prime factorization of a number? See:
http://magoosh.com/gmat/2012/gmat-math-factors/

When you know the prime factorization of a number, it's as if you know the DNA of the number. You have such profound knowledge about it.

For example, 35 = 5*7. That's the prime factorization. This means, for every factor of 35 we have, we need another factor of 7. There is absolutely no way we can make a factor of 35 without using a factor of 7 as one of the ingredients. If we only get three factors of 7, as is evident from the left side of the equation, that means we could make, at most, only three factors of 35. Once those three factors of 7 are used up, building the three factors of 35, then there is absolutely no possible of creating any more factors of 35, because we are out of one of the essential ingredients.

Does this make sense?
Mike
_________________

Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

SVP
Joined: 08 Jul 2010
Posts: 2017
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

03 Jun 2015, 06:27
roygush wrote:
If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both positive integers, how many different possible values of n are there?

A. 1
B. 2
C. 3
D. 4
E. 6

So what I did is basically making 35^n = (5^n)(7^n)

Thought the answer is B but i was wrong.
Can anyone explain?

TIP FOR SUCH QUESTIONS: All such Questions Require PRIME FACTORISATION i.e. Breaking the number into Prime factors and their power form on both sides of the equation

(3^4)(5^6)(7^3) = (35^n)(x)

i.e. (3^4)(5^6)(7^3) = (5*7)^n *(x)

i.e. (3^4)(5^6)(7^3) = (5^n)*(7^n) *(x)

This clearly depicts one thing which is x must be a multiple of (3^4) for sure and other factors of x will depend on the value of n

Please not that n is a positive integer

i.e. at n=1, (3^4)(5^6)(7^3) = (5^1)*(7^1) *(x) i.e. x=(3^4)(5^5)(7^2)
i.e. at n=2, (3^4)(5^6)(7^3) = (5^2)*(7^2) *(x) i.e. x=(3^4)(5^4)(7^1)
i.e. at n=3, (3^4)(5^6)(7^3) = (5^3)*(7^3) *(x) i.e. x=(3^4)(5^3)(7^0)

Only 3 possibilities

[Reveal] Spoiler:
C

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Retired Moderator
Joined: 18 Sep 2014
Posts: 1193
Location: India
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

01 Mar 2016, 11:31
If $$(3^4)(5^6)(7^3) = (35^n)(x)$$, where x and n are both positive integers, how many different possible values of n are there?

A. 1
B. 2
C. 3
D. 4
E. 6
_________________

The only time you can lose is when you give up. Try hard and you will suceed.
Thanks = Kudos. Kudos are appreciated

http://gmatclub.com/forum/rules-for-posting-in-verbal-gmat-forum-134642.html
When you post a question Pls. Provide its source & TAG your questions
Avoid posting from unreliable sources.

My posts
http://gmatclub.com/forum/beauty-of-coordinate-geometry-213760.html#p1649924
http://gmatclub.com/forum/calling-all-march-april-gmat-takers-who-want-to-cross-213154.html
http://gmatclub.com/forum/possessive-pronouns-200496.html
http://gmatclub.com/forum/double-negatives-206717.html
http://gmatclub.com/forum/the-greatest-integer-function-223595.html#p1721773

Math Expert
Joined: 02 Sep 2009
Posts: 44298
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

01 Mar 2016, 15:50
Nevernevergiveup wrote:
If $$(3^4)(5^6)(7^3) = (35^n)(x)$$, where x and n are both positive integers, how many different possible values of n are there?

A. 1
B. 2
C. 3
D. 4
E. 6

Merging topics. Please refer to the discussion above.
_________________
BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2494
GRE 1: 323 Q169 V154
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

06 Mar 2016, 21:12
roygush wrote:
If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both positive integers, how many different possible values of n are there?

A. 1
B. 2
C. 3
D. 4
E. 6

So what I did is basically making 35^n = (5^n)(7^n)

Thought the answer is B but i was wrong.
Can anyone explain?

here => 3^4 is immaterial
now values of n can be found from the following expressions
on LHS there can be (35)^1 or (35)^2 or (35)^3 but beyond that we dont have the excess power of 5 to utilize
thus => N=1,2,3 => 3 values.
_________________

Getting into HOLLYWOOD with an MBA

The MOST AFFORDABLE MBA programs!

STONECOLD's BRUTAL Mock Tests for GMAT-Quant(700+)

Average GRE Scores At The Top Business Schools!

Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4679
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

07 Mar 2016, 12:22
1
KUDOS
Expert's post
Chiragjordan wrote:
roygush wrote:
If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both positive integers, how many different possible values of n are there?

A. 1
B. 2
C. 3
D. 4
E. 6

So what I did is basically making 35^n = (5^n)(7^n)

Thought the answer is B but i was wrong.
Can anyone explain?

here => 3^4 is immaterial
now values of n can be found from the following expressions
on LHS there can be (35)^1 or (35)^2 or (35)^3 but beyond that we dont have the excess power of 5 to utilize
thus => N=1,2,3 => 3 values.

Dear Chiragjordan,
Yes, you are quite right. The 3^4 is entirely immaterial, and the factors of 7 limit the possibilities. This is a relatively straightforward problem. This problem inspired me to create a slightly more challenging problem:
if-a-and-b-are-positive-integers-and-147953.html
Enjoy!
Mike
_________________

Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

SVP
Joined: 12 Dec 2016
Posts: 1947
Location: United States
GMAT 1: 700 Q49 V33
GPA: 3.64
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]

### Show Tags

19 Apr 2017, 19:04
oh man, i miss the word positive, so i thought n can be o, so i chose D
Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit   [#permalink] 19 Apr 2017, 19:04
Display posts from previous: Sort by

# If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit

 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®.