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different factors

Expert replies
by madhur_ahuja » Fri Jul 17, 2009 11:30 pm
How many different factors does the integer n have ?

1. n=a^4b^3 , where a and b are different positive prime numbers.
2. The only positive prime numbers that are factors of n are 5 and 7 .
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Source: — Data Sufficiency |

by vittalgmat » Sat Jul 18, 2009 12:10 am
Hi Madhur,
Whenever u post a question , pls post the OA within the spoilers
That way we dont have to wait and can move on.


thanks
-V
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Re: different factors

by vittalgmat » Sat Jul 18, 2009 12:14 am
I got A.

here is my soln.

Stmt 1 says
n = a^4 *b^3 where a and b are diff prime numbers.

So total number of factors = (4+1)*(3+1) = 20.

sufficient.


Stmt 2:
The only positive prime numbers that are factors of n are 5 and 7 .


So we can write it as
n = 5^a * 7^b.
Since we dont know a and b, we cannot find out the total # of factors.

So not sufficient.

Hence A
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by adamsmith2009 » Sat Jul 18, 2009 7:58 pm
I got A.

here is my soln.

Stmt 1 says
n = a^4 *b^3 where a and b are diff prime numbers.

So total number of factors = (4+1)*(3+1) = 20.

sufficient.


Stmt 2:
The only positive prime numbers that are factors of n are 5 and 7 .


So we can write it as
n = 5^a * 7^b.
Since we dont know a and b, we cannot find out the total # of factors.

So not sufficient.

Hence A

Vittal,

Is there a standard formula for number of factors. You used (4+1)*(3+1) = 20. So is it simply (power of X+1)*(power of Y+1)=# of factors?

Also, with B it's saying that there are only 2 prime factors 5 and 7. To me, that means that there are no other factors since every other number can be simplified to a prime factor (ex, 6, 4, 8, 9) which is why I chose B. Is this correct?

Just wanted to point statement one actually says (1. n=a^4b^3 , where a and b are different positive prime numbers.) There should be a * between the powers because it reads like you are raising a to the 4 raised to b to the 3.
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by vittalgmat » Sat Jul 18, 2009 8:15 pm
Yes, there is a standard formula.
If you decompose a number N to its prime factors such that
N = a^i * b^j * c*k..., where a, b and c are prime numbers.
and i, j , k are +ve integers, then the total number of factors is
(i+1) * (j+1) *(k+1).


U are already wondering why we add 1 to the powers. This +1 accounts for 0th power of the corr prime number. ie a^0 =1 is also a factor of N.



in a^4b^3 in this problem the * is implicit.. Pls read it as
a^4 *b^3

Also, with B it's saying that there are only 2 prime factors 5 and 7. To me, that means that there are no other factors since every other number can be simplified to a prime factor (ex, 6, 4, 8, 9) which is why I chose B. Is this correct?


From B we only know the prime factors. From that we cannot predict all the factors. ie we need to know the powers of these prime factors to find the number of prime factors.
Put this another way: From B, we know that 5 and 7 are prime factors.
But we dont know if N = 5*7 or N=5^a * 7^b.
So this is insufficient.

HT Helps



Ht Helps
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by adamsmith2009 » Tue Jul 21, 2009 9:24 am
Thanks Vittal. That helps.
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by GID09 » Tue Jul 21, 2009 1:49 pm
Is the question stem not asking "how many different factors are there?" ...so both 1 & 2 are sufficient, right?

GID09
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by vittalgmat » Tue Jul 21, 2009 9:07 pm
GID09 wrote:Is the question stem not asking "how many different factors are there?" ...so both 1 & 2 are sufficient, right?

GID09
The question is asking "how many different factors are there"
NOT
"how many different PRIME factors are there"
So 2 is insufficient.
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