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Integer Properties

by topspin330 » Mon Dec 23, 2013 7:16 pm
Can someone please help solve the following problem:

If m = k · p, where k and m are different positive integers, then does m have more than 5 prime factors?

(1) k has 5 different prime factors.
(2) p has 5 different prime factors.

Thanks.
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by Brent@GMATPrepNow » Mon Dec 23, 2013 9:39 pm
topspin330 wrote:Can someone please help solve the following problem:

If m = k · p, where k and m are different positive integers, then does m have more than 5 prime factors?

(1) k has 5 different prime factors.
(2) p has 5 different prime factors.

Thanks.
Target question: Does m have more than 5 prime factors?

Given: m = k · p, where k and m are different positive integers

Let`s go straight to....

Statements 1 and 2 combined
There are several values of k and p that satisfy both statements. Here are two:
Case a: k = (2)(3)(5)(7)(11) and p = (2)(3)(5)(7)(13), in which case m = (2)(3)(5)(7)(11)(2)(3)(5)(7)(13), which means m has 6 prime factors (2, 3, 5, 7, 11, and 13), which means m has more than 5 prime factors
Case b: k = (2)(3)(5)(7)(11) and p = (2)(2)(3)(5)(7)(11), in which case m = (2)(3)(5)(7)(11)(2)(2)(3)(5)(7)(11), which means m has 5 prime factors (2, 3, 5, 7, and 11), which means m does NOT have more than 5 prime factors

Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer = E

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by theCodeToGMAT » Mon Dec 23, 2013 10:20 pm
topspin330 wrote: If m = k · p, where k and m are different positive integers, then does m have more than 5 prime factors?
Brent, I agree with your solution. However, I have a very basic doubt, aside from the posted question.

When we say "m have more than 5 prime factors" .. it can mean same prime factors as well? I mean if m were: 2x2x2x2x2x2 .. then we could have said Yes, "n" has more than 5 prime factors?
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by vipulgoyal » Mon Dec 23, 2013 11:13 pm
Hi Brent, with all due respect I disagree with the solution first of all, there is a differance b/w all prime no and all differant prime no
here
Case a: k = (2)(3)(5)(7)(11) and p = (2)(3)(5)(7)(13), in which case m = (2)(3)(5)(7)(11)(2)(3)(5)(7)(13), which means m has 6 prime factors (2, 3, 5, 7, 11, and 13), which means m has more than 5 prime factors
in this case m has 10 prime factors and 6 differant prime factors
now back to the qustion I think OA should be B, here is my explaination, please correct me if i am wrong
m and K are positive int
1. k= 2*3*5*7*11, p could be fraction or int, p=1/2 no, p = 13 yes, insuffi
2. p= 2*3*5*7*11, since k and m are positive int, k could be prime or not prime in either cases m must have more then 5 prime no, note that DS is not asking for differant prime no
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by Stuart@KaplanGMAT » Tue Dec 24, 2013 12:40 am
topspin330 wrote:Can someone please help solve the following problem:

If m = k · p, where k and m are different positive integers, then does m have more than 5 prime factors?

(1) k has 5 different prime factors.
(2) p has 5 different prime factors.

Thanks.
I'm going to almost agree with Vipulgoyal, but say that the answer should be "C", together.

Also, I'm assuming that the question was correctly transcribed and the "different" wasn't omitted from "does m have more than 5 prime factors"?

Since the Q stem omits the word "different" and the statements include that word, we have to treat the stem as implying that we count each prime factor once for each time it appears; in other words, if there are two 2s, that counts as 2 prime factors.

From (1), we know that k has 5 primes, but p could be a fraction and therefore m could have fewer primes than k.

For example, if k=2*3*5*7*11 and p=1/2, then the only prime factors of m would be 3, 5, 7 and 11.

From (2), we know that p has at least 5 primes and that k has to be an integer, but if k=1 then m=p and therefore m could only have 5 primes as well. (We know that k and m are different, but nothing says that m can't equal p!)

Together, we know that if k and p each have at least 5 prime factors, then m must have at least 10 prime factors, giving us a definite "yes". Choose C!
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by vipulgoyal » Tue Dec 24, 2013 1:35 am
you got it Stuart :D
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by Brent@GMATPrepNow » Tue Dec 24, 2013 7:13 am
theCodeToGMAT wrote:
topspin330 wrote: If m = k · p, where k and m are different positive integers, then does m have more than 5 prime factors?
Brent, I agree with your solution. However, I have a very basic doubt, aside from the posted question.

When we say "m have more than 5 prime factors" .. it can mean same prime factors as well? I mean if m were: 2x2x2x2x2x2 .. then we could have said Yes, "n" has more than 5 prime factors?
If m = (2)(2)(2)(2)(2) = 32, then I'd say that m has 1 prime factor.
My rationale is twofold.
First, the positive factors of 32 are: 1, 2, 4, 8, 16 and 32
One of the six factors is prime: 2
Five of the six factors are non-prime: 1, 4, 8, 16 and 32

Second, on page 237 of the OG12, the solution for a particular question reads,
7,150 = 2 × 5 × 5 × 11 × 13
Thus, 7,150 has four prime factors: 2, 5, 11, and 13.
Notice that we don't count the 5's twice.
Likewise, I wouldn't count each of the five 2's (in m) as a separate factor.

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by Brent@GMATPrepNow » Tue Dec 24, 2013 7:27 am
vipulgoyal wrote:Hi Brent, with all due respect I disagree with the solution first of all, there is a differance b/w all prime no and all differant prime no
here
Case a: k = (2)(3)(5)(7)(11) and p = (2)(3)(5)(7)(13), in which case m = (2)(3)(5)(7)(11)(2)(3)(5)(7)(13), which means m has 6 prime factors (2, 3, 5, 7, 11, and 13), which means m has more than 5 prime factors
in this case m has 10 prime factors and 6 differant prime factors
now back to the qustion I think OA should be B, here is my explaination, please correct me if i am wrong
m and K are positive int
1. k= 2*3*5*7*11, p could be fraction or int, p=1/2 no, p = 13 yes, insuffi
2. p= 2*3*5*7*11, since k and m are positive int, k could be prime or not prime in either cases m must have more then 5 prime no, note that DS is not asking for differant prime no
I think the keyword here is "factor"
Let's say that m = (2)(3)(5)(7)(11)(2)(3)(5)(7)(13)
Now let's start listing all of the positive factors of m.
We get: 1, 2, 3, 4, 5, 6, 7, 9, 10, ...
At this point, the prime factors will be those factors above that are also prime. They are 2, 3, 5, 7, etc.
If we start counting factors more than once, where do we stop?
We could say that 7 has an infinite number of factors, since 7 = (7)(1)(1)(1)(1)(1).....

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by vipulgoyal » Wed Jan 01, 2014 10:24 pm
Hi Brent,

I think that qustion in OG 12th ed P 237 asking no of prime factors in differant context, I am mentioning qustion for refrence below.
159. How many prime numbers between 1 and 100 are
factors of 7,150 ?
(A) One
(B) Two
(C) Three
(D) Four
(E) Five

here of course we shouldnt count 5 twice, but if we would have simply asked no of prime factors
of 8 the ans would be 3 (2x2x2), and the number of differant prime fctors will be only one which is 2

now coming back to that qriginal qustion with NO disrespect I still believe that OA should be C

Sturat any suggestions ???
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by Brent@GMATPrepNow » Thu Jan 02, 2014 6:46 am
vipulgoyal wrote:Hi Brent,

I think that qustion in OG 12th ed P 237 asking no of prime factors in differant context, I am mentioning qustion for refrence below.
159. How many prime numbers between 1 and 100 are
factors of 7,150 ?
(A) One
(B) Two
(C) Three
(D) Four
(E) Five

here of course we shouldnt count 5 twice, but if we would have simply asked no of prime factors
of 8 the ans would be 3 (2x2x2), and the number of differant prime fctors will be only one which is 2

now coming back to that qriginal qustion with NO disrespect I still believe that OA should be C

Sturat any suggestions ???
I understand what you're saying, but if we say that 8 has 3 prime factors (2, 2, and 2), then that seems to create other problems. For example, how many factors (prime and non-prime) does 8 have?

Cheers,
Brent
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