Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 01 Sep 2012
Posts: 122

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
Question Stats:
57% (00:59) correct 43% (01:17) wrong based on 567 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 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?
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
If my answer helped, dont forget KUDOS!
IMPOSSIBLE IS NOTHING




Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4667

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
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 RoygushI 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) Answer = CDoes 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
(3^4) (5^3) (5^3) (7^3) = (3^4) (5^3) (35^3) => n = 3, x = (3^4) (5^3) => answer C



Math Expert
Joined: 02 Sep 2009
Posts: 47084

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



Manager
Joined: 01 Sep 2012
Posts: 122

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: 4667

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: ifaandbarepositiveintegersand147953.htmlMike
_________________
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: 77
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.
Answer: C. 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: 47084

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.
Answer: C. 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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1837
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
Rearrange 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 Answer = 3 = C
_________________
Kindly press "+1 Kudos" to appreciate



Manager
Joined: 20 Dec 2013
Posts: 249
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: 4667

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
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/gmatmathfactors/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: 2120
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 integeri.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 Answer: Option
_________________
Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha email: info@GMATinsight.com I Call us : +919999687183 / 9891333772 Online OneonOne 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



BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2647

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.
_________________
MBA Financing: INDIAN PUBLIC BANKS vs PRODIGY FINANCE! Getting into HOLLYWOOD with an MBA! The MOST AFFORDABLE MBA programs!STONECOLD's BRUTAL Mock Tests for GMATQuant(700+)AVERAGE GRE Scores At The Top Business Schools!



Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4667

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
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: ifaandbarepositiveintegersand147953.htmlEnjoy! 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: 1874
Location: United States
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



Manager
Joined: 09 Mar 2017
Posts: 59
Location: India
GPA: 4
WE: Marketing (Advertising and PR)

Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]
Show Tags
03 May 2018, 21:46
Hi ! If in the case the value of n =4 ; we still get a value which +ve decimal.. Why cant we say then it has more values



Math Expert
Joined: 02 Sep 2009
Posts: 47084

Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]
Show Tags
03 May 2018, 23:26



Manager
Joined: 09 Mar 2017
Posts: 59
Location: India
GPA: 4
WE: Marketing (Advertising and PR)

Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit [#permalink]
Show Tags
03 May 2018, 23:28
Oh ! Thanks Bunuel




Re: If (3^4)(5^6)(7^3) = (35^n)(x), where x and n are both posit
[#permalink]
03 May 2018, 23:28






