To elaborate on the previous post, the problem is testing a very important concept which is to to find out the # of factors of a positive integer.
This concept is very often tested in GMAT in different ways. Given below is the concept explained in more details and some sample problems.
Concept:
A positive integer n which has prime factorization a^x * b^y * c^z has (x+1)*(y+1)*(z+1) factors. (note a, b and c are distinct prime factors of n).
eg1: 36 = 2^2 * 3^2
The # of positive factors of 36 = (2+1) * (2+1) = 9
Note that the # of factors include 1 and 36 as well.
eg2: 360 = 2^3 * 3^2 * 5
# of factors = (3+1) * (2+1) * (1+1) = 4 * 3 * 2 = 24
explanation: to obtain a factor of 360, from the above list, I can pick a 2 or a 3 or a 5 or any combination of 2,3,5. (think of this as a probability or a combination problem).
The # of 2's I can select are (one 2 or two 2's or three 2's or zero 2's)
The # of 3's I can select are (one 3 or two 3's or zero 3's)
The # of 5's I can select are (one 5 or zero 5's).
Note that select zero 2's, zero3's and zero 5's gives you the trivial case of 1 as a factor.
The total # of combinations = 4*3*2=24
Applications:
Application 1:
What are the Number of positive factors for 1800?
1800 = 18 * 100 = 2^3 * 3^2 * 5^2
# of factors = 4 * 3 * 3 = 36
Application 2:
How many positive integers k are there such that 100k is a factor of (2^2)(3) (5^3)?
100 = 4 *25 = 2^2 * 5^2
Given 100 k is a factor of (2^2) (3) (5^3)
Which means, 2^2 * 5^2 * k is a factor of (2^2) (3) (5^3)
You see that k can be equal to 1, 3, or 5 or 15. (note that positive integer means 1 also).
So, there are four integers 1, 3, 5 and 15 that would satisfy the given condition.
In other words, the problem translates to finding the # of factors for 15.
15 = 3 * 5
and hence, the # of factors = (1+1) * (1+1) = 2 * 2 = 4.
3. Approximately what percent of the positive factors of 1800 are less than 30?
(A) 22% (B) 28% (C) 32% (D) 38% (E) 42%
First we need to determine the # of factors for 1800.
In order to find out the # of factors, express 1800 = 18*100 = 2^3 * 3^2 * 5^2
The # of factors = (3+1) * (2+1) * (2+1) = 4 * 3 * 3 = 36
The factors that are less than 30 are 1,2,3,4,5,6,8,9,10,12,15,18,20,24,25 (note that 1 is included but not 30).
% = 15/36 * 100 = 42%
4. What can you tell about an integer n that has only (and only) 3 factors.
a. Note that 3 itself is a prime number with only 1 and 3 as the factors.
So, the number n has the prime factorization of n = p^2 where p is a prime number.
b. From the above result, one can see that n is a perfect sqaure.
for example, n = 4, 9, 25, 49, 121 etc
c. Note that 4 satisfies the above condition and hence, in general we cannot conclude that n is odd. However, if it is given that n>4, we safely conclude that n is odd.
d. Similarly, if a positive integer n has only 5 factors, what can you tell about n.
n can be written as p^4 where p is prime. Hence, you can conclude that sqrt(n) = p^2 is a perfect sqaure.
-mathguru
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