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
Prime Factors
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- logitech
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IMO it is Hlachlanc 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
LGTCH
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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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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)
P^2*t= P^3.P^(2t-3)=m*P^(2t-3)
Therefore m must be a multiple of P^2*t
(B)
Drill baby drill !
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- logitech
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Drill Bidisha Drill!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)
LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"
@cramya,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
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
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
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.