PS: IF the prime numbers p and t

[cesar.rodriguez.blanco]

I missed this question. Please help!

If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of (p^2)*t?

1) m has more than 9 positive factors
2) m is a multiple of p^3 [moderator's edit]

[aditinig]

What is this OA?

[agha79]

Is OA for this "B"

[nitin_prakash_khanna]

Statement 2 is wrong, it should say...

2> m is a multiple of p^3.

St 1 tells us that there are total > 9 factors for m, which is possible with even
m = (some integer N) * p * t^4.
So NOT Sufficient

St 2 tells us that
m = np^3 and we know that p & t are the prime factors of m
so its really
m = n * p^3 * t (raised to some unknown power)
So m = some int N * p^2 * t and hence Sufficient.

[RonPurewal]

I missed this question. Please help!

If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of (p^2)*t?

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

we already know that m is a multiple of t, so the only real issue here is whether there are 2 copies of "p" in its prime factorization.
i.e., we already know it's a multiple of pt; all that's missing is the second "p".

(1)
this clearly could be a "yes", if m is something like (p^1000)(t^1000). therefore, the challenge lies in looking for a "no".
we can get a "no" by keeping only one "p", and just raising "t" to a huge power. for instance, m = (p)(t^1000) will have over two thousand factors.
insufficient.

(2)
if m is a multiple of p^3, then it's at least a multiple of (p^3)(t), so, sufficient.

ans = (b)

[sachin.w]

I missed this question. Please help!

If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of (p^2)*t?

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

we already know that m is a multiple of t, so the only real issue here is whether there are 2 copies of "p" in its prime factorization.
i.e., we already know it's a multiple of pt; all that's missing is the second "p".

(1)
this clearly could be a "yes", if m is something like (p^1000)(t^1000). therefore, the challenge lies in looking for a "no".
we can get a "no" by keeping only one "p", and just raising "t" to a huge power. for instance, m = (p)(t^1000) will have over two thousand factors.
insufficient.

(2)
if m is a multiple of p^3, then it's at least a multiple of (p^3)(t), so, sufficient.

ans = (b)

Hi Ron,
I don't quite understand how second statement is sufficient.
Please explain with a numerical example..

[jlucero]

Hi Ron,
I don't quite understand how second statement is sufficient.
Please explain with a numerical example..

p is a positive prime number, so p could be 2, 3, 5, 7...
If m is a multiple of p^3, it must be divisible by 8, 27, 125, 343...

The question asks if m is divisible by p^2, so if the number is divisible by p^3, it must be divisible by p^2.

If p is 2
m is divisible by 8
so m must be divisible by 4

If p is 3
m is divisible by 27
so m must be divisible by 9

etc.

[sachin.w]

Hi Ron,
I don't quite understand how second statement is sufficient.
Please explain with a numerical example..

p is a positive prime number, so p could be 2, 3, 5, 7...
If m is a multiple of p^3, it must be divisible by 8, 27, 125, 343...

The question asks if m is divisible by p^2, so if the number is divisible by p^3, it must be divisible by p^2.

If p is 2
m is divisible by 8
so m must be divisible by 4

If p is 3
m is divisible by 27
so m must be divisible by 9

etc.

Thanks Joe, but we are asked if m a multiple of (p^2)*t

[RonPurewal]

Thanks Joe, but we are asked if m a multiple of (p^2)*t

read the thread, please. thanks.

quoting what i wrote above:

we already know that m is a multiple of t, so the only real issue here is whether there are 2 copies of "p" in its prime factorization.

[sachin.w]

Thanks Ron,

Just to cement my understanding..

Say 'r' is also a prime factor of m..

So, will 'm' be a multiple of 'pr'.

[jnelson0612]

Thanks Ron,

Just to cement my understanding..

Say 'r' is also a prime factor of m..

So, will 'm' be a multiple of 'pr'.

Yes, if we are assuming that the original information in the question stem is still in force--that p and t are prime factors of m.