GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Jul 2018, 14:37

### 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

# If the prime factorization of the integer q can be expressed

Author Message
TAGS:

### Hide Tags

Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 511
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
If the prime factorization of the integer q can be expressed [#permalink]

### Show Tags

Updated on: 18 Apr 2013, 21:38
9
35
00:00

Difficulty:

95% (hard)

Question Stats:

42% (02:04) correct 58% (02:13) wrong based on 589 sessions

### HideShow timer Statistics

If the prime factorization of the integer q can be expressed as $$a^{2x} b^{x} c^{3x-1}$$, where a, b, c, and x are distinct positive integers, which of the following could be the total number of factors of q?

(A) 3j + 4, where j is a positive integer
(B) 5k + 5, where k is a positive integer
(C) 6l + 2, where l is a positive integer
(D) 9m + 7, where m is a positive integer
(E) 10n + 1, where n is a positive integer

Guys - does this question makes sense to you? I am struggling. OA is not provided either.

_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Originally posted by enigma123 on 28 Jan 2012, 17:12.
Last edited by walker on 18 Apr 2013, 21:38, edited 2 times in total.
Math Expert
Joined: 02 Sep 2009
Posts: 47037

### Show Tags

28 Jan 2012, 17:27
20
15
enigma123 wrote:
If the prime factorization of the integer q can be expressed as a^2x b^x c^3x-1, where a, b, c, and x are distinct positive integers, which of the following could be the total number of factors of q?
(A) 3j + 4, where j is a positive integer
(B) 5k + 5, where k is a positive integer
(C) 6l + 2, where l is a positive integer
(D) 9m + 7, where m is a positive integer
(E) 10n + 1, where n is a positive integer

Guys - does this question makes sense to you? I am struggling. OA is not provided either.

Given: $$q=a^{2x}*b^x*c^{3x-1}$$. # of distinct factors of q is $$(2x+1)*(x+1)*(3x-1+1)=(2x+1)*(x+1)*3x$$.

Now, $$3x(x+1)(2x+1)$$ is both multiple of 3 and even (as either x or x+1 is even). Let's check the answer choices:

A. 3j+4 --> not a multiple of 3 (1 more than multiple of 3) --> discard;
B. 5k+5 --> can be even and multiple of 3, for example if k=5, then 5k+5=30 --> keep;
C. 6l+2 --> not a multiple of 3 (2 more than multiple of 3) --> discard;
D. 9m+7 --> not a multiple of 3 (1 more than multiple of 3) --> discard;
E. 10n+1 --> not even (10n+1=even+odd=odd) --> discard.

So, only 5k+5 can possibly be the total number of factors of q (in fact the lowest value of k would be 17 for which x=2).

Hope it's clear.

P.S. In case one doesn't know.
Finding the Number of Factors of an Integer

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
For more on number properties check: math-number-theory-88376.html
_________________
##### General Discussion
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 511
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)

### Show Tags

28 Jan 2012, 17:28
1
Are you a Maths Avatar or something? You are indeed a GMATCLUB LEGEND. Thanks buddy.
_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Current Student
Joined: 20 Mar 2014
Posts: 2641
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: If the prime factorization of the integer q can be expressed [#permalink]

### Show Tags

11 Jul 2015, 13:38
2
1
enigma123 wrote:
If the prime factorization of the integer q can be expressed as $$a^{2x} b^{x} c^{3x-1}$$, where a, b, c, and x are distinct positive integers, which of the following could be the total number of factors of q?

(A) 3j + 4, where j is a positive integer
(B) 5k + 5, where k is a positive integer
(C) 6l + 2, where l is a positive integer
(D) 9m + 7, where m is a positive integer
(E) 10n + 1, where n is a positive integer

Guys - does this question makes sense to you? I am struggling. OA is not provided either.

Straight forward question if you know the formula for number of factors of an integer $$q= a^x * b^y * c^z$$ , where $$x,y,z \geq{0}$$ and are integers = (x+1)(y+1)(z+1) including 1 and integer q.

Now given, q= $$a^{2x} b^{x} c^{3x-1}$$, thus based on the above formula, the number of factors of q including 1 and q = (2x+1)(x+1)(3x-1+1) = 3x(2x+1)(x+1). So we know that the number will be of the form 3x(2x+1)(x+1) , will be an integer and will be true for ALL X. Thus substitute x=1, we get number of factors = 3*3*2 =18 (this eliminates A, as 3 is a factor of 18 and thus can not leave 4 or (4-3=1) as the remainder. Similarly, eliminate C and D).

Now substitute x=2, we get number of factors = 6*5*3 = 90 and this eliminates E as 10 is a factor of 90 and will leave a remainder of 0. Thus B is the only remaining answer choice and is the OA.
Senior Manager
Joined: 03 Apr 2013
Posts: 290
Location: India
Concentration: Marketing, Finance
Schools: Simon '20
GMAT 1: 740 Q50 V41
GPA: 3
Re: If the prime factorization of the integer q can be expressed [#permalink]

### Show Tags

20 Jun 2016, 17:57
1
IMO (B)
The answer can be thought of in a simple manner. Here's my take..

First, we know that calculating the number of factors by formula will transform into something like

$$(2x+1)(x+1)(3x)$$

when this is simplified..it becomes

$$6x^3 + 9x^2 + 3x$$

This means two things

1. The number is a multiple of 3.
2. The value of this number depends on the value of x(as x can be taken out as common).

Looking at the options
we see that option B is $$5k + 5$$ or $$5(k+1)$$ or simply a multiple of 5.
The question asks "could be"..so the value of x could be 5..right?
and there you have it
_________________

Spread some love..Like = +1 Kudos

Intern
Joined: 07 Feb 2017
Posts: 11
If the prime factorization of the integer q can be expressed [#permalink]

### Show Tags

30 Jun 2018, 18:41
Bunuel wrote:
enigma123 wrote:
If the prime factorization of the integer q can be expressed as a^2x b^x c^3x-1, where a, b, c, and x are distinct positive integers, which of the following could be the total number of factors of q?
(A) 3j + 4, where j is a positive integer
(B) 5k + 5, where k is a positive integer
(C) 6l + 2, where l is a positive integer
(D) 9m + 7, where m is a positive integer
(E) 10n + 1, where n is a positive integer

Guys - does this question makes sense to you? I am struggling. OA is not provided either.

Given: $$q=a^{2x}*b^x*c^{3x-1}$$. # of distinct factors of q is $$(2x+1)*(x+1)*(3x-1+1)=(2x+1)*(x+1)*3x$$.

Now, $$3x(x+1)(2x+1)$$ is both multiple of 3 and even (as either x or x+1 is even). Let's check the answer choices:

A. 3j+4 --> not a multiple of 3 (1 more than multiple of 3) --> discard;
B. 5k+5 --> can be even and multiple of 3, for example if k=5, then 5k+5=30 --> keep;
C. 6l+2 --> not a multiple of 3 (2 more than multiple of 3) --> discard;
D. 9m+7 --> not a multiple of 3 (1 more than multiple of 3) --> discard;
E. 10n+1 --> not even (10n+1=even+odd=odd) --> discard.

So, only 5k+5 can possibly be the total number of factors of q (in fact the lowest value of k would be 17 for which x=2).

Hope it's clear.

P.S. In case one doesn't know.
Finding the Number of Factors of an Integer

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
For more on number properties check: http://gmatclub.com/forum/math-number-theory-88376.html

Hey Bunuel,

solving for total number of factors, $$q=6*x^3+9*x^2+3*x$$

so q will be always even. But in choice b, if k is even, then q will be odd. How can this be resolved ?
Math Expert
Joined: 02 Sep 2009
Posts: 47037
Re: If the prime factorization of the integer q can be expressed [#permalink]

### Show Tags

30 Jun 2018, 23:47
Dineshrambalaji wrote:
Bunuel wrote:
enigma123 wrote:
If the prime factorization of the integer q can be expressed as a^2x b^x c^3x-1, where a, b, c, and x are distinct positive integers, which of the following could be the total number of factors of q?
(A) 3j + 4, where j is a positive integer
(B) 5k + 5, where k is a positive integer
(C) 6l + 2, where l is a positive integer
(D) 9m + 7, where m is a positive integer
(E) 10n + 1, where n is a positive integer

Guys - does this question makes sense to you? I am struggling. OA is not provided either.

Given: $$q=a^{2x}*b^x*c^{3x-1}$$. # of distinct factors of q is $$(2x+1)*(x+1)*(3x-1+1)=(2x+1)*(x+1)*3x$$.

Now, $$3x(x+1)(2x+1)$$ is both multiple of 3 and even (as either x or x+1 is even). Let's check the answer choices:

A. 3j+4 --> not a multiple of 3 (1 more than multiple of 3) --> discard;
B. 5k+5 --> can be even and multiple of 3, for example if k=5, then 5k+5=30 --> keep;
C. 6l+2 --> not a multiple of 3 (2 more than multiple of 3) --> discard;
D. 9m+7 --> not a multiple of 3 (1 more than multiple of 3) --> discard;
E. 10n+1 --> not even (10n+1=even+odd=odd) --> discard.

So, only 5k+5 can possibly be the total number of factors of q (in fact the lowest value of k would be 17 for which x=2).

Hope it's clear.

P.S. In case one doesn't know.
Finding the Number of Factors of an Integer

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.
For more on number properties check: http://gmatclub.com/forum/math-number-theory-88376.html

Hey Bunuel,

solving for total number of factors, $$q=6*x^3+9*x^2+3*x$$

so q will be always even. But in choice b, if k is even, then q will be odd. How can this be resolved ?

The question asks: which of the following COULD be the total number of factors of q?

Options A, C, D and E, CANNOT be the total number of factors of q, regardless of the values of j, l, m, and n.

Option B, 5k + 5, COULD be the total number of factors of q, for specific values of k. For example if k = 17.
_________________
Re: If the prime factorization of the integer q can be expressed   [#permalink] 30 Jun 2018, 23:47
Display posts from previous: Sort by

# Events & Promotions

 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®.