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Prime Factors

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by lachlanc » Sat Nov 08, 2008 12:50 pm
If the prime numbers p and t are the only prime factors of integer m, is m a mulitple of p^2*t?

1) m has more than 9 positive factors
2) m is a multiple of p^3
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Re: Prime Factors

by logitech » Sat Nov 08, 2008 12:56 pm
lachlanc wrote:If the prime numbers p and t are the only prime factors of integer m, is m a mulitple of p^2*t?

1) m has more than 9 positive factors
2) m is a multiple of p^3
IMO it is H
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by cramya » Sat Nov 08, 2008 1:28 pm
If the prime numbers p and t are the only prime factors of integer m, is m a mulitple of p^2*t?

1) m has more than 9 positive factors
2) m is a multiple of p^3

It would be B)


Q: Is m a multiple of p^2 t

Given: p and t are the only prime factors of m

Stmt I

FYI:
We can obtain m's factors by adding 1 to the exponents of p and t and multiplying them

Eg: No of factors of 6 are four(1,2,3 and 6)


Express 6 as the product of its prime factors

6= 2 ^ 1 * 3 ^1 (dropping the bases ; adding 1 to the 2's and 3's exponents and multiplying together)

(1+1) (1+1) = 4


m has more 9 positive factors. This could be p ^ 1 t ^ 9 Still m would have more than 9 positive factors. p^2 t is not a multiple of m

or it could be p^ 2 t ^ 3 Then m is a multiple of p^2 t

We dont know for sure INSUFF

Stmt II

m is a multiple of p^3 so m contains p ^ 3 in it somewhere(if not m cannot be a multiple of p^3). It is also given t is a prime factor therefore m will be a multiple of p^2 t

SUFF

B)

Let me know if u still have questions
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by logitech » Sat Nov 08, 2008 1:44 pm
Cramya, please help me to understand the statement II. I am still not clear about where the t is in P^3
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by cramya » Sat Nov 08, 2008 1:48 pm
t is not in p^ 3 ; t is in the given info (t is also a prime factor of m)
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by Bidisha800 » Sat Nov 08, 2008 9:45 pm
m is multiple of P^3

P^2*t= P^3.P^(2t-3)=m*P^(2t-3)

Therefore m must be a multiple of P^2*t

(B)
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by logitech » Sat Nov 08, 2008 9:49 pm
Bidisha800 wrote:m is multiple of P^3

P^2*t= P^3.P^(2t-3)=m*P^(2t-3)

Therefore m must be a multiple of P^2*t

(B)
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by nitin86 » Thu Nov 13, 2008 9:10 am
cramya wrote:
If the prime numbers p and t are the only prime factors of integer m, is m a mulitple of p^2*t?

1) m has more than 9 positive factors
2) m is a multiple of p^3

It would be B)


Q: Is m a multiple of p^2 t

Given: p and t are the only prime factors of m

Stmt I

FYI:
We can obtain m's factors by adding 1 to the exponents of p and t and multiplying them

Eg: No of factors of 6 are four(1,2,3 and 6)


Express 6 as the product of its prime factors

6= 2 ^ 1 * 3 ^1 (dropping the bases ; adding 1 to the 2's and 3's exponents and multiplying together)

(1+1) (1+1) = 4


m has more 9 positive factors. This could be p ^ 1 t ^ 9 Still m would have more than 9 positive factors. p^2 t is not a multiple of m

or it could be p^ 2 t ^ 3 Then m is a multiple of p^2 t

We dont know for sure INSUFF

Stmt II

m is a multiple of p^3 so m contains p ^ 3 in it somewhere(if not m cannot be a multiple of p^3). It is also given t is a prime factor therefore m will be a multiple of p^2 t

SUFF

B)

Let me know if u still have questions
@cramya,

I think answer should be D.
from the logic that you have provided for calculating the factors of any number.

So, the logic was, if a number M has only 2 prime factors , we can write it as M = (P)^a * (T)^b, where P and T are prime factor and a and b are powers of these prime factor respectively.

Now, Total number of positive factors of M can be calculated as
(a+1) * (b+1),
[This is by the same logic, as used by @cramya to calculate the factors of 6]

Now, if according to STMT 1, M has 9 positive factors, then that means
Total number of factors of M = 9 = (a+1) * (b+1)

Now, (a+1) * (b+1) = 9, means that various values that (a+1) and (b+1) can take to make 9 are

1) 9 * 1
2) 1 * 9
3) 3 * 3

but if any of (a+1) or (b+1) takes value as 1, then that mean that mean either a or b has to be equal to ZERO.

But according to the question, a and b can't be zero, as both P and T are prime factors of M.

Hence, (a+1) and (b+1) both are equal to 3, or each of a and b is equal to 2

Hence M = (P)^2 * (T)^2

Hope this helps
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by mental » Thu Nov 13, 2008 9:18 am
the question says....MORE THAN 9 FACTORS

your solution is for exactly 9 factors

(a+1)(b+1) > 9

a=1, b=4........one of the possibilities
a=2, b=3........another

I is insufficient
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by stop@800 » Thu Nov 13, 2008 9:34 am
Incorrect: Please ignore
I assumed Option A to be
"It has 9 factors".

============

Avoid so much calculations.

Lets look at what the question says:

m is multiple of p and t
we do not to what powers but each exists at least once.

We need to find is m multiple of p^2*t
that means do we have an extra p or not


A
9 factors
9 can only be derived by 3*3 [using product of integers]
hence m has to be of the form p^2 t^2
hence extra p is in m
so Suff

B
multiple of p^3
we just needed an extra p
but here we got two :)
Sufficient

Ans IMO is D


The catch here is:
People will commit mistake by not dividing 9 into factors.


cramya,
p ^ 1 t ^ 9 will have
(1+1)(9+1) = 20 factors

remeber the a,b rule

p1^a p2^b will have
(a+1)(b+1) factors
Last edited by stop@800 on Thu Nov 13, 2008 10:40 am, edited 1 time in total.
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by lachlanc » Thu Nov 13, 2008 10:31 am
OA is B. You the man, cramya.
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