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If the prime numbers p and t are the only prime factors of

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Re: If the prime numbers p and t are the only prime factors of [#permalink]

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New post 13 Feb 2017, 18:11
If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of t*p^2?
(1) m has more than 9 positive factors
(2) m is a multiple of p^3

Draw a prime box and put p and t inside. According to the problem, there could be multiple instances of p and t in there, but that's it. We want to know whether there are at least two p's and one t in there.

Start with statement 2. If m is a multiple of p^3, that means there are 3 p's in m's prime box. There's already a t in there, according to the original question. So there are at least 2 p's and one t. Answer to question is yes, so statement is sufficient. Eliminate A, C, E.

Statement 1. Notice that this just says "positive factors" NOT prime factors. The complete set of factors is made by multiplying the prime factors in different combinations. For example, 12 has the prime factors 2, 2, and 3. We can find all of the general factors of 12 by taking 2, 3, 2*2, 2*3, 2*2*3, and of course 1.

So m has more than 9 positive factors. Well, I know m has p and t - there are 2 factors. And I know m has 1 and itself - there are 2 more factors, for a total of four. I need five more, so I have to add to my prime box to be able to create five more general factors. The only things I can put in my prime box are p and t. I can put all p's, all t's, or some combination of p's and t's. If I put in at least one p, then I'd have at least 2 p's and one t, which would answer the question "yes." BUT, if I put in all t's, then I'd only have one p, which would answer the question "no" - so the statement is insufficient.

Hence B.
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Re: If the prime numbers p and t are the only prime factors of [#permalink]

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New post 14 Oct 2017, 03:35
Hi there,

Small request - to avoid confusion can we please update the last part of the question to be "is m a multiple of (p^2)*t" rather than p^2*t?

Thanks!

phoenixgmat wrote:
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

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Re: If the prime numbers p and t are the only prime factors of [#permalink]

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New post 14 Oct 2017, 05:24
acyn wrote:
Hi there,

Small request - to avoid confusion can we please update the last part of the question to be "is m a multiple of (p^2)*t" rather than p^2*t?

Thanks!

phoenixgmat wrote:
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


You should be familiar how math notation works. p^2*t can ONLY mean (p^2)*(t), nothing else. If it were p^(2*t), then it would have been written that way.
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If the prime numbers p and t are the only prime factors of [#permalink]

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New post 14 Oct 2017, 19:55
phoenixgmat wrote:
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

Given: m is a multiple of p*t and p,t are the only prime factors of m

Statement 1: Not enough info. Cannot say if m is a multiple of p*p*t. Not sufficient

Statement 2: m is a multiple of p*p*p. Sufficient.

Hence B.
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Re: If the prime numbers p and t are the only prime factors of [#permalink]

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New post 17 Oct 2017, 19:04
Bunuel wrote:
phoenixgmat wrote:
I would appreciate some help with:

If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p²t?
1) m has more than 9 positive factors.
2) m is a multiple of p³

some explanations to both statements would be great!
thx a lot


We are told that \(p\) and \(t\) are the ONLY prime factors of m. It could be expressed as \(m=p^x*t^y\), where \(x\) and \(y\) are integers \(\geq{1}\).

Question: is \(m\) a multiple of \(p^2*t\). We already know that \(p\) and \(t\) are the factors of \(m\), so basically question asks whether the power of \(p\), in our prime factorization denoted as \(x\), more than or equal to 2: so is \(x\geq{2}\).

(1) m has more than 9 positive factors:

Formula for counting the number of distinct factors of integer \(x\) expressed by prime factorization as: \(n=a^x*b^y*c^z\), is \((x+1)(y+1)(z+1)\). This also includes the factors 1 and \(n\) itself.

We are told that \((x+1)(y+1)>9\) (as we know that \(m\) is expressed as \(m=p^x*t^y\))
But it's not sufficient to determine whether \(x\geq{2}\). (\(x\) can be 1 and \(y\geq{4}\) and we would have their product \(>9\), e.g. \((1+1)(4+1)=10\).) Not sufficient.

(2) m is a multiple of p^3
This statement clearly gives us the value of power of \(p\), which is 3, \(x=3>2\). So \(m\) is a multiple of \(p^2t\). Sufficient.

Answer: B.


Hi Bunuel,

This statement clearly gives us the value of power of p, which is 3, x=3>2

How???

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Re: If the prime numbers p and t are the only prime factors of [#permalink]

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New post 17 Oct 2017, 20:14
zanaik89 wrote:
Bunuel wrote:
phoenixgmat wrote:
I would appreciate some help with:

If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p²t?
1) m has more than 9 positive factors.
2) m is a multiple of p³

some explanations to both statements would be great!
thx a lot


We are told that \(p\) and \(t\) are the ONLY prime factors of m. It could be expressed as \(m=p^x*t^y\), where \(x\) and \(y\) are integers \(\geq{1}\).

Question: is \(m\) a multiple of \(p^2*t\). We already know that \(p\) and \(t\) are the factors of \(m\), so basically question asks whether the power of \(p\), in our prime factorization denoted as \(x\), more than or equal to 2: so is \(x\geq{2}\).

(1) m has more than 9 positive factors:

Formula for counting the number of distinct factors of integer \(x\) expressed by prime factorization as: \(n=a^x*b^y*c^z\), is \((x+1)(y+1)(z+1)\). This also includes the factors 1 and \(n\) itself.

We are told that \((x+1)(y+1)>9\) (as we know that \(m\) is expressed as \(m=p^x*t^y\))
But it's not sufficient to determine whether \(x\geq{2}\). (\(x\) can be 1 and \(y\geq{4}\) and we would have their product \(>9\), e.g. \((1+1)(4+1)=10\).) Not sufficient.

(2) m is a multiple of p^3
This statement clearly gives us the value of power of \(p\), which is 3, \(x=3>2\). So \(m\) is a multiple of \(p^2t\). Sufficient.

Answer: B.


Hi Bunuel,

This statement clearly gives us the value of power of p, which is 3, x=3>2

How???


x is a power of p. m is a multiple of p^3, so x > 3.
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Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If the prime numbers p and t are the only prime factors of   [#permalink] 17 Oct 2017, 20:14

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