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np - prime numbers

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by gaurvipul » Mon Oct 15, 2012 3:54 am
Acc. to me
(E) p^3 and np

Explanation:

Consider n^4
As n is prime, n^4 is a perfect square. By rule, perfect squares will always have odd number of factors.

Consider p^3, substituting a prime number will help us solve it fast, e.g. 3^3
Factor pair will give 4 factors >> [(3^3,3^1);(3^2,3^2)]

Consider np, substituting prime numbers will help us solve it fast, e.g. n=2 & p=3

4 factors >> [(2,3),(6,1)]
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by The Iceman » Mon Oct 15, 2012 8:17 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 key to solving this one is to understand that any Natural number can be formed as a decomposition of prime factors.

Every natural number N can be expressed as N = P1^a * P2^b * P3^c * P4^d .... so on and so forth...

where P1, P2,... are prime numbers and a,b,c are powers of the respective prime numbers.

The number of positive divisors of P1^a = a+1, because the factors vary from a=0 to a=a; there are a+1 in number :).

The number of positive factors of N can be found out by the multiplication counting principle. The number of positive factors of N = (a+1)(b+1)(c+1)(d+1)...

The key thing to understand here is that we are only calculating the number of +ve factors. If we are to find total number of factors, simply multiple the above expression by 2 as the negative factors will be symmetrically placed and '0' can never be a factor by definition.

The question here precisely talks about the number of +ve divisors, so we will not multiply by 2.

This concept can be used on other similar problems to get to the result.
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by The Iceman » Mon Oct 15, 2012 8:19 pm
The Iceman wrote:
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 key to solving this one is to understand that any Natural number can be formed as a decomposition of prime factors.

Every natural number N can be expressed as N = P1^a * P2^b * P3^c * P4^d .... so on and so forth...

where P1, P2,... are prime numbers and a,b,c are powers of the respective prime numbers.

The number of positive divisors of P1^a = a+1, because the factors vary corresponding to the range from a=0 to a=a; these are a+1 in number :).

The number of positive factors of N can be found out by the multiplication counting principle. The number of positive factors of N = (a+1)(b+1)(c+1)(d+1)...

The key thing to understand here is that we are only calculating the number of +ve factors. If we are to find total number of factors, simply multiple the above expression by 2 as the negative factors will be symmetrically placed and '0' can never be a factor by definition.

The question here precisely talks about the number of +ve divisors, so we will not multiply by 2.

This concept can be used on other similar problems to get to the result.
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by ritind » Thu Dec 27, 2012 12:50 am
@sairam : Although number crunching makes this ques really simple

Here's a faster approach, if u hv less time up ur sleeve
It is mentioned that n and p are prime numbers, then np is definitely goin to be one of the ans
coz the product will have following factors 1, n, p and np
So u r left with only 3 choices i.e. C, D and E

If a number is quadrapled then it is bound to have more than 4 factors. Factors will be 1, p, p^2, p^3 and p^4

So nw u r left with C and E. 50% of getting a question right without even touchin pen or paper
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by rajeshsinghgmat » Fri Feb 08, 2013 12:31 am
(E) p^3 and np

Let, p =2, n =3, np =2*3

the divisiors or p^3 are, 1, 2,4 and 8.

the divisors of np are, 1,2,3 and 6.
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by ritzzzr » Tue Feb 12, 2013 9:43 pm
we know prime numbers have two divisors 1 & the number itself
if we square it we have three divisors : 1 the number itself & its square
eg if number is 3 then 3^2 will have three divisors 1,3 & 9
so we have 4 divisors for n^3.
When we multiply two prime numbers eg
3 & 7
we have 1 3 7 & 21 similarly for others .

SO we have 4 divisors for n^3 & np.
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by smvjkumar » Tue Apr 23, 2013 10:56 pm
the answer is E

Let us assume 2 prime no
n = 2 and p = 3

now n^4 = 2^4 = 16 -> has 5 divisors

p^3 = 3^3 = 27 -> has 4 divisors

np = 2*3 = 6 -> 4 divisors

hence p^3 and np have 4 divisors
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by Java_85 » Mon Sep 02, 2013 7:39 am
I'd Say E ,
np has the 1,n,p and np
p3 has the 1,p,pp and ppp
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by Mathsbuddy » Mon Dec 09, 2013 7:38 am
n^4 -> 1, n, n^2, n^3, n^4 -> 5 divisors
p^3 -> 1, p, p^2, p^3 -> 4 divisors
np -> 1, n, p, np -> 4 divisors

ANSWER = (E) p^3 and np
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by lulufrenchie » Tue Sep 23, 2014 7:59 am
I got E by picking n= 2 and p=3 and then I tried again with n=3 and p = 2.
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by [email protected] » Tue Sep 23, 2014 10:12 am
Hi lulufrenchie,

As you've come to realize, this question is perfect for TESTing Values. That strategic approach can help you to solve lots of questions in the Quant section on Test Day.

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by lulufrenchie » Wed Sep 24, 2014 6:24 am
[email protected] wrote:Hi lulufrenchie,

As you've come to realize, this question is perfect for TESTing Values. That strategic approach can help you to solve lots of questions in the Quant section on Test Day.

GMAT assassins aren't born, they're made,
Rich
Hi Rich,

Did I lose time by testing two values for each one of them ?
How can you recognize when you need to test it once or several times ?

Best,
Lulu
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by [email protected] » Wed Sep 24, 2014 9:00 am
Hi lulufrenchie,

For this question, just one TEST is required, so to answer your question - you spent a bit more time than you needed to to get the correct answer. That having been said, if a little extra work will GUARANTEE that you get the correct answer, then there's some logic to spending that little bit of extra time. The danger comes from spending WAY TOO MUCH time (staring at the screen, for example) on any one question - that would NOT be a good use of your time.

Usually, on PS questions, just one TEST is required. There are exceptions though: the phrase "MUST be true" means that you might have to TEST a second or third option. When doing a TEST, if more than 1 answer is correct, then at least one extra TEST would be necessary to find the 1 answer that is always correct (and eliminate whatever answer(s) is "sometimes correct). When using this approach on DS questions, at least 2 TESTs are required (and sometimes more than that).

As you continue to practice this tactic, you'll start to get a better sense of which questions require more work. Keep an eye out for patterns in the wording of the question - those phrases will tell you what you need to do.

GMAT assassins aren't born, they're made,
Rich
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by jaspreetsra » Wed Oct 22, 2014 4:11 pm
E making sense.
My explanation:
Let n = 2, p= 3,
np = 2*3 =6
So
2^4; divisors: 2^1, 2^2, 2^3, 2^4, and 1
3^3; divisors: 3^1, 3^2, 3^3, and 1
6; divisors: 2, 3, 6, and 1
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by nikhilgmat31 » Mon Jun 22, 2015 2:34 am
I think question should explicity tell that 4 distinct factors.

lets p=3
3^3 leads 3*3*3 which can have factors as - 1,3,3,3,9,27 which is not 4 which leads to Answer C.

What's say of others.
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