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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?

Re: Prime factor:Can someone give a definite answer? [#permalink]
22 Jun 2010, 02:06

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, 04:51, edited 1 time in total.

Re: Prime factor:Can someone give a definite answer? [#permalink]
22 Jun 2010, 02:11

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.

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?

Re: If m is divisible by 3, how many prime factors does m have? [#permalink]
22 Oct 2013, 22:55

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.

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