It is currently 16 Jan 2018, 11:41

### 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 is the product of the 5 different prime numbers, how many facto

Author Message
TAGS:

### Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4669

Kudos [?]: 3302 [0], given: 0

GPA: 3.82
If n is the product of the 5 different prime numbers, how many facto [#permalink]

### Show Tags

28 Jun 2016, 17:40
Expert's post
6
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

56% (00:39) correct 44% (01:31) wrong based on 91 sessions

### HideShow timer Statistics

If n is the product of the 5 different prime numbers, how many factors n have except 1 and n?
A. 20
B. 22
C. 24
D. 30
E. 32

*An answer will be posted in 2 days.
[Reveal] Spoiler: OA

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Find a 10% off coupon code for GMAT Club members.
“Receive 5 Math Questions & Solutions Daily”
Unlimited Access to over 120 free video lessons - try it yourself

Kudos [?]: 3302 [0], given: 0

Manager
Joined: 22 Jun 2016
Posts: 244

Kudos [?]: 115 [1], given: 10

Re: If n is the product of the 5 different prime numbers, how many facto [#permalink]

### Show Tags

28 Jun 2016, 20:48
1
KUDOS
1
This post was
BOOKMARKED
Number of all factors of a number:

if N=(a^m)*(b^n)*(c^o)*(d^p)... (where a,b,c,d.. are prime factors of N)
so, number of all factors of a N = (m+1)*(n+1)*(o+1)*(p+1)....

Applying the above in the question:
We have a no. N which has 5 prime factors.

So, we can say N = (a^m)*(b^n)*(c^o)*(d^p)*(e^q) (where a,b,c,d and e are 5 prime factor of N)

Lowest possible N will have m=n=o=p=q=1.

No. of factors of such N = (m+1)*(n+1)*(o+1)*(p+1)*(q+1) = 2*2*2*2*2 = 32

Now, just remove 1 and N from 32 and we are left with 30 factors.

_________________

P.S. Don't forget to give Kudos

Kudos [?]: 115 [1], given: 10

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4669

Kudos [?]: 3302 [0], given: 0

GPA: 3.82
Re: If n is the product of the 5 different prime numbers, how many facto [#permalink]

### Show Tags

01 Jul 2016, 01:17
Expert's post
1
This post was
BOOKMARKED
(1+1)(1+1)(1+1)(1+1)(1+1)=32. Hence, from 32-2=30, we can see that D is the correct answer.
_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Find a 10% off coupon code for GMAT Club members.
“Receive 5 Math Questions & Solutions Daily”
Unlimited Access to over 120 free video lessons - try it yourself

Kudos [?]: 3302 [0], given: 0

Senior Manager
Joined: 20 Feb 2015
Posts: 387

Kudos [?]: 112 [0], given: 10

Concentration: Strategy, General Management
If n is the product of the 5 different prime numbers, how many facto [#permalink]

### Show Tags

01 Jul 2016, 02:13
1
This post was
BOOKMARKED
n=2*3*5*7*11
no of factors of n = 2*2*2*2*2=32 (take any other (set of) distinct prime number for n and the result will always be the same)
no of factors except 1 and n = 32 -2 =30

Kudos [?]: 112 [0], given: 10

VP
Joined: 22 May 2016
Posts: 1244

Kudos [?]: 454 [0], given: 670

If n is the product of the 5 different prime numbers, how many facto [#permalink]

### Show Tags

17 Dec 2017, 14:24
MathRevolution wrote:
If n is the product of the 5 different prime numbers, how many factors n have except 1 and n?
A. 20
B. 22
C. 24
D. 30
E. 32

*An answer will be posted in 2 days.

If the method does not look familiar*:
To find the number of factors in a number, n, including n and 1:

1) Find n's prime factors raised to the proper power. Here they are given.
This n could be ANY five distinct primes

$$2^15^111^113^131^1$$

2) Take each prime factor's power and add 1 to the exponent.
E.g., factor 2 has a power of 1. 1+1 = 2. Do the same for all factors.

2 2 2 2 2

3) Multiply the resultant numbers

$$2^5 = 32$$ factors of n, including n and 1

The question excludes n and 1, so without those two factors, there are
(32 - 2) = 30 factors

*See Bunuel , Finding the Number of Factors of an Integer
_________________

(formerly genxer123)
At the still point, there the dance is. -- T.S. Eliot

Kudos [?]: 454 [0], given: 670

If n is the product of the 5 different prime numbers, how many facto   [#permalink] 17 Dec 2017, 14:24
Display posts from previous: Sort by