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question on GMAT Prep exam

by oberlove » Wed Oct 20, 2010 7:16 am
I came across the problem below on one of the GMAT Prep exams. Can someone help me figure out how to solve this problem? In general, I find these type of questions a real bear. Any general strategies in approaching them would also be appreciated. Thanks!

If the prime numbers 'p' and 't' are the only prime factors of the integer 'm', is 'm' a multiple of p^2*t?
(i) 'm' has more than 9 positive factors
(ii) 'm' is a multiple of p^3
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Source: — Data Sufficiency |

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by Rahul@gurome » Wed Oct 20, 2010 7:52 am
Given: The prime numbers 'p' and 't' are the only prime factors of the integer 'm'.
=> m = (p^a)*(t^b) , where a and b are greater than or equal to 1. In simple words, m is composed of one or more p's and one or more t's.

Now one possible case is m = pt. In this case, m is not a multiple of p^2*t. Therefore we need more information.

Statement 1: m has more than 9 positive factors.
=> m is composed of more than one p's and/or more than one t's.

Some possible cases are m = p*t^k , where k is greater than or equal to 4. In these cases m has more than 9 positive factors.

Like for k = 4, m = p*t^4. The factors are 1, p, t, t^2, t^3, t^4, pt, p*t^2, p*t^3 and p*t^4.

But in these cases, m is not a multiple of p^2*t.

Not SUFFICIENT.

Statement 2: m is multiple of p^3.
=> m = (p^3)^a*t^b , where a and b are greater than or equal to 1. In simple words, m is composed of three or more p's and one or more t's.

Thus m always contains p^2*t in it. Because, m is multiple p^3 => m is multiple of p^2 and m has t as a prime factor => m is a multiple of t. Thus, m is multiple of p^2*t.

SUFFICIENT.

The correct answer is B.
Last edited by Rahul@gurome on Wed Oct 20, 2010 8:11 am, edited 1 time in total.
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by shovan85 » Wed Oct 20, 2010 7:54 am
There is a formula you should know of:

N = a^x*b^y*c^z where a b c are distinct prime numbers and x y and z are corresponding powers then
Total number of factors = (x+1)*(y+1)*(z+1)

first break the number to all possible prime numbers (along with the powers) Then add one to each power and multiple.

Try the same for 36 = 2^2 * 3^2
Total factors = (2+1)*(2+1) = 9 which are [1,2,3,4,6,9,12,18,36]

Now,

m has only two PRIME factors so we can have other factors but those will not be prime.

1. m has more than 9 positive factors:

so there is no guarantee that the m will be a multiple of (p^2)*t.

For example if I take m = p*(t^5) then total number of factors of m = (1+1)*(5+1) = 12 which is > 9.

So not sufficient.

2. m is a multiple of p^3.

so we can write m = (p^3)* (t^x) ; x in the sense any integer > 0 as t has to be a factor of m (mentioned in the Q)

Clearly P^2 is a factor of p^3
[As p^3 can be written as p*(p^2)]

Thus Sufficient.

IMO
B
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