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Prime Factors

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by danjuma » Sun Oct 10, 2010 9:18 am
If the prime numbers p and t are the only prime factors of th integer m, is m a multiple of p square times t?

1. M has more than 9 positive factors

2. M is a multiple of P to the power 3
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Source: — Data Sufficiency |
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by shovan85 » Sun Oct 10, 2010 9:47 am
IMO B
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by clock60 » Sun Oct 10, 2010 10:27 am
here i got E but not too sure, need verification and oa
if i got the problem right
m=p^k*t^n, where p and t are both primes, and exponents k, n integers
we need to find is
(p^k*t^n)/(p^2*t)=integer. ( from problem i confused with is m a multiple of p square times t?) what is p-square??

(1) m has more that 9 factors. posssible numbers
m=p^3*t^2 ( the number of factors 4*3=12) and this m is a multiple of p^2*t. but on the other hand
m=p^1*t^4 (the number of factors 2*5=10) but m is not for sure the multiple of p^2*t
so 1 st insuff

(2)m=p^k*t^(3b), i mean m=p^k*t^3. or m=p^k*t^6. for sure m is divisible by t, but we can`t say that it is divisible by p^2, also insuff

both also insuff
m=p^1*t^6 ( number of factors 2*7=14) and it is not divisible by p^2*t.
m=p^2*t^3 (number od factors 3*4=12) and it is divisible by p^2*t

hope i did not miss something
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by shovan85 » Sun Oct 10, 2010 11:28 am
clock60 wrote:here i got E but not too sure, need verification and oa
if i got the problem right
m=p^k*t^n, where p and t are both primes, and exponents k, n integers
we need to find is
(p^k*t^n)/(p^2*t)=integer. ( from problem i confused with is m a multiple of p square times t?) what is p-square??

(1) m has more that 9 factors. posssible numbers
m=p^3*t^2 ( the number of factors 4*3=12) and this m is a multiple of p^2*t. but on the other hand
m=p^1*t^4 (the number of factors 2*5=10) but m is not for sure the multiple of p^2*t
so 1 st insuff

(2)m=p^k*t^(3b), i mean m=p^k*t^3. or m=p^k*t^6. for sure m is divisible by t, but we can`t say that it is divisible by p^2, also insuff

both also insuff
m=p^1*t^6 ( number of factors 2*7=14) and it is not divisible by p^2*t.
m=p^2*t^3 (number od factors 3*4=12) and it is divisible by p^2*t

hope i did not miss something
2. M is a multiple of P to the power 3

I think it means M=(p^3)*(t^k)

If M is multiple of p^3 then it must be a multiple of p^2.
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by clock60 » Sun Oct 10, 2010 11:43 am
yes friend you are right, my silly mistake and thanks for pointing out!
(2) m=p^k*t^n, k>=3, here p^k will always be multiple of p^2. as k>=3, and as n>=1, t^n will always be multiple of t(p^3/p^2)*(t^n/t)=p*t^(n-1)=integer
the answer is B
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by shovan85 » Sun Oct 10, 2010 12:09 pm
clock60 wrote:yes friend you are right, my silly mistake and thanks for pointing out!
(2) m=p^k*t^n, k>=3, here p^k will always be multiple of p^2. as k>=3, and as n>=1, t^n will always be multiple of t(p^3/p^2)*(t^n/t)=p*t^(n-1)=integer
the answer is B
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