Author 
Message 
Current Student
Joined: 11 May 2008
Posts: 556

confusing ONE [#permalink]
Show Tags
08 Aug 2008, 09:59



Director
Joined: 10 Sep 2007
Posts: 938

Re: confusing ONE [#permalink]
Show Tags
08 Aug 2008, 10:10
1
This post received KUDOS
Question says M = p^xt^y. Given Atleast x=1, and y=1. Question is asking is x>=2, Already given t is a factor so we are not bother about it.
1) Says m has 9 factor may be x=1 and t=8, or x=8 and t=1. This is may be case so question cannot be answered and A, D out.
2) Tells us x=3. So question is answered. Answer B.



Current Student
Joined: 11 May 2008
Posts: 556

Re: confusing ONE [#permalink]
Show Tags
08 Aug 2008, 10:15
well, does'nt look like it is asking that.. it asks if m is divisible by p^2*t. i dont think that is the same thing as is power of p>=2.... right??? or.... wrong what it looks to be asking is when m/p^2*t, do we get an integer??? but i think that if p and t divide m , then p^2 usually divides right, unless the m itself <p^2*t. eg , take 6. it is divisible by 2 and 3 , but not by 2^2*3, but that is bcos p^2*t>m for this example... but what is the problem asking us?



Senior Manager
Joined: 23 May 2006
Posts: 324

Re: confusing ONE [#permalink]
Show Tags
08 Aug 2008, 10:19
Good explanation.



Director
Joined: 10 Sep 2007
Posts: 938

Re: confusing ONE [#permalink]
Show Tags
08 Aug 2008, 10:20
arjtryarjtry wrote: well, does'nt look like it is asking that.. it asks if m is divisible by p^2*t. i dont think that is the same thing as is power of p>=2.... right??? or.... wrong I have rephrased the question. Thereby it looks different. If M is divisible by p^2t then of course it means that m has factor of p equal to or more than 2 and also that m has a factor of t.



SVP
Joined: 29 Aug 2007
Posts: 2473

Re: confusing ONE [#permalink]
Show Tags
08 Aug 2008, 10:25
arjtryarjtry wrote: 1: to have more than 9 +ve factors, either, at least, p = 4 and t = 1, or p = 1 and t = 4. nsf. 2: m is a multiple of p^3 means m is also a multiple of p^2t. suff So B.
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



SVP
Joined: 29 Aug 2007
Posts: 2473

Re: confusing ONE [#permalink]
Show Tags
08 Aug 2008, 10:27
abhijit_sen wrote: arjtryarjtry wrote: well, does'nt look like it is asking that.. it asks if m is divisible by p^2*t. i dont think that is the same thing as is power of p>=2.... right??? or.... wrong I have rephrased the question. Thereby it looks different. If M is divisible by p^2t then of course it means that m has factor of p equal to or more than 2 and also that m has a factor of t. That could be a question, but GMAT questions donot have that pattern. it is: is m/[(p^2)(t)] = k?
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



Director
Joined: 27 May 2008
Posts: 543

Re: confusing ONE [#permalink]
Show Tags
08 Aug 2008, 10:39
one important formula to know
if m can be written as m = p^x * q^y * r ^z where p, q and r are distinct prime numbers number for total +ve factors = (1+x)(1+y)(1+z)
question : m = p^x * t^y is m a mutiple of p^2 * t > is x >=2
statement 1 : (1+x)(1+y) > 9 .... x,y = 1,4 Or 3,3 Not suff statement 2 : m = k * p^3 > x >=3 .. Suff



Current Student
Joined: 11 May 2008
Posts: 556

Re: confusing ONE [#permalink]
Show Tags
08 Aug 2008, 10:49
durgesh79 wrote: one important formula to know
if m can be written as m = p^x * q^y * r ^z where p, q and r are distinct prime numbers number for total +ve factors = (1+x)(1+y)(1+z)
question : m = p^x * t^y is m a mutiple of p^2 * t > is x >=2
statement 1 : (1+x)(1+y) > 9 .... x,y = 1,4 Or 3,3 Not suff statement 2 : m = k * p^3 > x >=3 .. Suff why this combination?? how did u arrive at that combo of 1,4 or 3,3? why not 2,5 or something like that?



Director
Joined: 27 May 2008
Posts: 543

Re: confusing ONE [#permalink]
Show Tags
08 Aug 2008, 10:53
arjtryarjtry wrote: durgesh79 wrote: one important formula to know
if m can be written as m = p^x * q^y * r ^z where p, q and r are distinct prime numbers number for total +ve factors = (1+x)(1+y)(1+z)
question : m = p^x * t^y is m a mutiple of p^2 * t > is x >=2
statement 1 : (1+x)(1+y) > 9 .... x,y = 1,4 Or 3,3 Not suff statement 2 : m = k * p^3 > x >=3 .. Suff why this combination?? how did u arrive at that combo of 1,4 or 3,3? why not 2,5 or something like that? we have to see if (1+x)(1+y) > 9 is enough to prove that x >= 2 or not , so i took two examples with x < 2 and x > 2. Since both cases are possible, the statement is not suff.



VP
Joined: 17 Jun 2008
Posts: 1381

Re: confusing ONE [#permalink]
Show Tags
08 Aug 2008, 11:08
arjtryarjtry wrote: given that : Only two prime factors are p,t of m question : m=p^2 * t *n n is any integer n is again product of p,t only since p,t are only prime factors (1) says m has > 9 positive factors => t^8.p or p^8.t or t^2*p^7 etc hence when m=p*t^8 then its not the multiple of p^2 *t when m = p^8.t then its the multiple of p^2 *t INSUFFI (2) says m is multiple of p^3 also only two prime factors of m are p,t hence p^3 * p^k * t^l is possible hence always multiple of p^2*t SUFFI IMO B
_________________
cheers Its Now Or Never



Intern
Joined: 06 Aug 2008
Posts: 2

Re: confusing ONE [#permalink]
Show Tags
08 Aug 2008, 20:36
I will first take the example and then i will explain it... Say m = 36. So m has 9 factors: 1 2 3 4 6 9 12 18 36. Now only two numbers are prime in this: 2 and 3. say p and t. so p^2*t^2 = m. so, A is definately the answer, because p^2 *t is factor of m, obviously.
Further, in general, we have (a + 1)(b+1)(c+1) factors of a number where a, b, c are powers of prime number in the factors of number....
see, example, 36 has 9 factors....(3)(3)=9. so a = 2, b = 2. because there are only two primes 2, 3.
So clearly A is the answer. Not B.



Senior Manager
Joined: 06 Apr 2008
Posts: 435

Re: confusing ONE [#permalink]
Show Tags
08 Aug 2008, 23:38
arjtryarjtry wrote: It is B) Suppose there is a number 12, now it has only two prime factors 2 and 3 Statement 1) is insuff. because you can have \(3^9*2\) and it is not multiple of \(2^2 * 3\) Statement 2) if m is multiple of \(p^3\) then it will be multiple of \(p^2\) as well and t is already another prime factor so multiple of \(p^2 * t\)










