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# If m is divisible by 3, how many prime factors does m have?

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If m is divisible by 3, how many prime factors does m have? [#permalink]

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22 Jun 2010, 00:12
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If m is divisible by 3, how many prime factors does m have?

(1) $$\frac{m}{3}$$ is divisible by 3

(2) $$\frac{m}{3}$$ has two different prime factors

[Reveal] Spoiler:
I've seen answers given as B , C and E by different people.
I would go for E
1. if m has 3 and m/3 has a 3, then m has at least 3^2, but m may have other factors..
2.m has a 3 and 2 different primes (x,y) from m/3. So m has at least 3,x,y if not more. But either x or y could be a 3 as well. in that case m has 3^2,x or 3^2,y
1+2
m/3 has a 3 and another prime factor x
m then has 3^2, x, and other factors that we dont know about

According to mgmat book, since m is a variable, we couldn't tell how many prime factors if we don't have constraint m
Could someone post the correct solution?

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-m-is-divisible-by-3-how-many-prime-factors-does-m-have-126571.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Oct 2013, 23:53, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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22 Jun 2010, 03:06
Pipp wrote:
According to mgmat book, since m is a variable, we couldn't tell how many prime factors if we don't have constraint m

what does this mean ?

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Last edited by intellijat on 22 Jun 2010, 05:51, edited 1 time in total.
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22 Jun 2010, 03:11
m is divisible by 3. How many prime factors does m have?

1. Remember, we also consider zero to be perfectly divisible by any number, and hence m could be some multiple of 3 greater than 3, or zero. INSUFF

2. m/3 has 2 prime factors. say x and y. So m = 3 * x^k * y^n
If m = 2*3*5, then m/3 = 2*5. So m has 3 prime factors
If m = 3^2 * 5 then m/3 = 3*5, but m has only 2 prime factors, insufficient.

Both 1 and 2.

Since m/3 is divisible by 3, one of the prime factors of m/3 is 3
So m/3 = 3^n * y^k => m = 3^(n+1) * y^k
Thus m also has 2 prime factors.

Pick C.

Remember we are only concerned with prime factors, which are counted only once each irrespective of how many times they occur in the prime factorization of a number.
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Re: If m is divisible by 3, how many prime factors does m have? [#permalink]

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22 Oct 2013, 16:11
Pipp wrote:
If m is divisible by 3, how many prime factors does m have?
1). m/3 is divisible by 3.
2). m/3 has two different prime factors.

[Reveal] Spoiler:
B , C and E
by different people.
I would go for E
1. if m has 3 and m/3 has a 3, then m has at least 3^2, but m may have other factors..
2.m has a 3 and 2 different primes (x,y) from m/3. So m has at least 3,x,y if not more. But either x or y could be a 3 as well. in that case m has 3^2,x or 3^2,y
1+2
m/3 has a 3 and another prime factor x
m then has 3^2, x, and other factors that we dont know about

According to mgmat book, since m is a variable, we couldn't tell how many prime factors if we don't have constraint m
Could someone post the correct solution?

Hi Bunuel

Can you explain this question

Regards
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Re: If m is divisible by 3, how many prime factors does m have? [#permalink]

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22 Oct 2013, 23:55
Expert's post
If m is divisible by 3, how many prime factors does m have?

(1) $$\frac{m}{3}$$ is divisible by 3 --> $$\frac{m}{3}=3k$$ --> $$m=3^2*k$$ --> $$m$$ has at least one prime 3, but it can have more than one, in case $$k$$ has some number of other primes. Not sufficient

2) $$\frac{m}{3}$$ has two different prime factors --> first of all 3 is a factor of $$m$$, so 3 is one of the primes of $$m$$ for sure.

Now, if power of 3 in $$m$$ is more than or equal to 2 then $$m$$ will have have only two prime factors: 3 and one other, example: $$m=18$$, (as in $$\frac{m}{3}$$ one 3 will be reduced, at least one more 3 will be left, plus one other, to make the # of different factors of $$\frac{m}{3}$$ equal to two. Thus $$m$$ will have 3 and some other prime as a prime factors).

But if $$m$$ has 3 in power of one then $$m$$ will have 3 prime factors: 3 and two others, example $$m=30$$ (one 3 will be reduced in $$m$$ and $$\frac{m}{3}$$ will have some other two prime factors, which naturally will be the primes of $$m$$ as well). Not sufficient.

(1)+(2) From (1) $$3^2$$ is a factor of $$m$$, thus from (2) $$m$$ has only two distinct prime factors: 3 and one other. Sufficient.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-m-is-divisible-by-3-how-many-prime-factors-does-m-have-126571.html
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Re: If m is divisible by 3, how many prime factors does m have?   [#permalink] 22 Oct 2013, 23:55
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