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# confusing ONE

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08 Aug 2008, 09:59
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

guys ...i just didnt get this problem.... pls can some one tell me what exactly is being asked?? and how to start analysis of such problems... also what difficulty level is this problem at? on a scale of 1 to 5? 5 being most difficult...

and what is the meaning of sentence 1?? it says m has 9 +ve factors. in the stem it says m has 2 prime factors... how can a no. have factors like this??
is it that square and cube of the basic prime nos are also considered??
how do we start thinking abt such problems on the exam day???

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Director
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08 Aug 2008, 10:10
1
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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.

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Current Student
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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?

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08 Aug 2008, 10:19
Good explanation.

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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.

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08 Aug 2008, 10:25
arjtryarjtry wrote:
guys ...i just didnt get this problem.... pls can some one tell me what exactly is being asked?? and how to start analysis of such problems... also what difficulty level is this problem at? on a scale of 1 to 5? 5 being most difficult...

and what is the meaning of sentence 1?? it says m has 9 +ve factors. in the stem it says m has 2 prime factors... how can a no. have factors like this??
is it that square and cube of the basic prime nos are also considered??
how do we start thinking abt such problems on the exam day???

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.
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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?
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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

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Current Student
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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?

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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.

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08 Aug 2008, 11:08
arjtryarjtry wrote:
guys ...i just didnt get this problem.... pls can some one tell me what exactly is being asked?? and how to start analysis of such problems... also what difficulty level is this problem at? on a scale of 1 to 5? 5 being most difficult...

and what is the meaning of sentence 1?? it says m has 9 +ve factors. in the stem it says m has 2 prime factors... how can a no. have factors like this??
is it that square and cube of the basic prime nos are also considered??
how do we start thinking abt such problems on the exam day???

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
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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.

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08 Aug 2008, 23:38
arjtryarjtry wrote:
guys ...i just didnt get this problem.... pls can some one tell me what exactly is being asked?? and how to start analysis of such problems... also what difficulty level is this problem at? on a scale of 1 to 5? 5 being most difficult...

and what is the meaning of sentence 1?? it says m has 9 +ve factors. in the stem it says m has 2 prime factors... how can a no. have factors like this??
is it that square and cube of the basic prime nos are also considered??
how do we start thinking abt such problems on the exam day???

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 $$p^2 * t$$

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Re: confusing ONE   [#permalink] 08 Aug 2008, 23:38
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