BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

np - prime numbers

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 39
Joined: Thu Feb 25, 2010 7:04 am
Thanked: 2 times

np - prime numbers

by sairamGmat » Wed Aug 18, 2010 6:20 am
If n and p are different positive prime numbers, which of the integers n^4, p^3 and np has (have) exactly 4 positive divisors?
(A) n^4 only
(B) p^3 only
(C) np only
(D) n^4 and np
(E) p^3 and np
Top
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 324
Joined: Mon Jul 05, 2010 6:44 am
Location: London
Thanked: 70 times
Followed by:3 members

by kmittal82 » Wed Aug 18, 2010 6:43 am
Factors of n^ 4 :

n,n^2,n^3,n^4,1 = 5 divisors/factors

Factors of p^3:

p,p^2,p^3,1 = 4 divisors/factors

Factors of np:

n,p,np,1 = 4 divisors/factors



Hence, (E) should be the answer. Can you pls confirm?
Top
User avatar
Legendary Member
Posts: 1460
Joined: Tue Dec 29, 2009 1:28 am
Thanked: 135 times
Followed by:7 members

by selango » Wed Aug 18, 2010 6:59 am
If a and b are primes such that a^m and b^n then number of factors of a and b can be expressed as (m+1) and (n+1)

a^m.b^n=(m+1)*(n+1)

n and p are primes and have exactly 4 positive divisors.

This means power should be 3 or 2 terms raise to power of 1.

n^3 has (3+1) or 4 factors.

np= n^1.p^1=2.2=4 factors

Pick E
--Anand--
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Wed Aug 18, 2010 1:09 pm
sairamGmat wrote:If n and p are different positive prime numbers, which of the integers n^4, p^3 and np has (have) exactly 4 positive divisors?
(A) n^4 only
(B) p^3 only
(C) np only
(D) n^4 and np
(E) p^3 and np
The algebraic solutions are great, but almost certainly the quickest way to solve is by picking numbers.

Just let n=2 and p=3:

n^4 = 2^4 = 16... the factors of 16 are 1, 2, 4, 8, 16... not exactly 4, so eliminate (A) and (D).

p^3 = 3^3 = 27... the factors of 27 are 1, 3, 9, 27... exactly 4, so eliminate (C).

np = 2*3 = 6... the factors of 6 are 1, 2, 3 and 6... exactly 4, so choose (E)!
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Fri Sep 24, 2010 10:39 am
GMAT Score:650

by sumit88 » Fri Jul 15, 2011 7:05 am
Any prime number has only 2 divisors.

If we square a prime no., number of divisors will always be greater than 2. eg 2 has 2 divisors(1 and 2), but 2^2 has 3 divisors(1,2,4).

Similarly as we increase the power(exponent) divisors will increase. eg 2^5 = 32 have (1,2,4,8,16,32).

Thus when two primes are multiplied it will lead to exactly 4 divisors(1, prime1, prime2, product).

Anyways its always better to key-in numbers.
Take Risks In Your Life If YoU Win, YoU Can Lead ...!! If YoU Loose, YoU CaN Guide ~ Swami Vivekananda
Top
Master | Next Rank: 500 Posts
Posts: 401
Joined: Tue May 24, 2011 1:14 am
Thanked: 37 times
Followed by:5 members

by MBA.Aspirant » Fri Jul 15, 2011 7:46 am
p^3 has 4 factors

np

say n= 3 and p =2

6 has 1,2,3, 6 (4 factors)
Top
Senior | Next Rank: 100 Posts
Posts: 72
Joined: Wed Oct 27, 2010 2:06 am
Thanked: 2 times
Followed by:1 members

Formula behind factors

by saurabh2525_gupta » Sun Aug 07, 2011 8:29 pm
The answer is E.

The rule is:- For distinct prime number a,b,c

Let there is a number k = a^p * b^q * c*r
The number of divisors of number k = (p+1)(q+1)(r+1).


Best Regards,
JOHN
Top
User avatar
Junior | Next Rank: 30 Posts
Posts: 27
Joined: Tue Aug 02, 2011 2:50 am

by wingsoffire » Mon Sep 05, 2011 8:49 am
I 'll Mark E.
Top
User avatar
Junior | Next Rank: 30 Posts
Posts: 27
Joined: Tue Aug 02, 2011 2:50 am

by wingsoffire » Mon Sep 05, 2011 8:50 am
I 'll Mark E.
Top
Legendary Member
Posts: 608
Joined: Sun Jun 19, 2011 11:16 am
Thanked: 37 times
Followed by:8 members

by saketk » Mon Sep 12, 2011 12:03 pm
option E
Top
User avatar
Newbie | Next Rank: 10 Posts
Posts: 7
Joined: Sat Aug 13, 2011 6:04 am
Thanked: 2 times
GMAT Score:690

by PhoenixId » Sat Oct 01, 2011 5:34 am
Since n and p are positive prime numbers, the product np has 4 factors.
Apart from 1 and itself, n and p are also factors of np.
So total - 1, itself (product np), n and p.

p^3 will have 4 factors - apart from 1 and itself, p and p^2 are factors.

n^4 will have 5 factors - 1, itself, n^3, n^2 and n.

My pick - E
- Retaking GMAT, aiming for 740
Top
Master | Next Rank: 500 Posts
Posts: 160
Joined: Tue Jul 07, 2009 1:09 pm
Thanked: 1 times
Followed by:1 members

by Sharma_Gaurav » Mon Jan 09, 2012 2:10 pm
straight E.
Top
User avatar
Legendary Member
Posts: 626
Joined: Fri Dec 23, 2011 2:50 am
Location: Ahmedabad
Thanked: 31 times
Followed by:10 members

by ronnie1985 » Fri Feb 03, 2012 10:21 am
p^3 and pn
Follow your passion, Success as perceived by others shall follow you
Top
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Thu Apr 19, 2012 9:34 pm

by nerdyguy » Mon May 21, 2012 11:37 am
Ans is (E)..
p^3 has divisors: 1, p, p^2 and p^3
np has divisors: 1, n, p, np

n^4 has 5 divisors
Top
User avatar
Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Thu Sep 06, 2012 8:06 am

by [email protected] » Sat Oct 06, 2012 11:38 pm
I have marked E.
np is a sure answer for any prime number s it is true.
and p^4 cant be the case as power greater than 3 will not give exactly 4 divisors.
So we are left with E.
Top
Post new topic Post Reply