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A positive odd number n has exactly 8 factors. How many prime factors does it have?

(1) n is less than 150 (2) n is greater than 100

Agree with MA ...and how to find those numbers not by blind guessing
Employ the formula of number of factors:
Assume that n is constituted by prime factors a,b,c,etc ... the number of factors of n is the product of those (factors' powers + 1).
Observe that 8= 1*8 or 2*4 or 2*2*2 , in which 1*8 is impractical coz 1 can't be (factor's power+1) --> we have two cases: 2*4 or 2*2*2. That is to say, n can have 2 prime factors or 3 prime factors. Since n is an odd number, n can't have factor of 2. Now, look at statement 1 and 2, we can estimate the biggest prime factor that n can have is 7 . Thus we have two cases: n has prime factor 3, 5 and 7 .
The case with 3 prime factors fits the case of 2*2*2 --> each prime factor has power of 1 --> n= 3*5*7= 105
The case with 2 prime factors fits the case of 2*4 --> one prime factor has power of 1, the other 3 ..try 3^3 *5 , we have n= 135

==> we can point out at least 2 cases of n which lie in the open range of either statement 1 or 2 or both statements.

--> E.

It's just my thought of how to look at this problem from another angle but it's not really advisable in real test

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.

How stunned I am to know that you were an Olympiad winner in maths . I'm sure GMATclubbers will benefit a lot if you often chime in so please be around

8 = 8*1 or 4*2 or 2*2*2
Case1 when we take 8 = 8*1
Case2 when we take 8 = 4*2
Case3 when we take 8 = 2*2*2
Total factors of a number x^a * y^b * z^c are (a+1) (b+1) (c+1)
St1:
Case1: n = x^7 where x is a prime number. Taking least odd prime n goes beyod 150. Reject
Case2: n = x^3 * y^1. Only values of x and y are x = 3 and y =5 for n to be less than 150. Then n = 135
Case3: n = x^1 * y^1 * z^1. Only values of x,y and z are x = 3 y = 5 and z = 7 for n to be less than 150. So n = 105
So n may be 105 or 135: INSUFF

St2: Clearly INSUFF. in all three cases above x,y and z could be any prime number greater than 2.

Dahiya - well done and thanks for clarifying the application of the concept explained above. Precisely. Only three possibilities if a number has only 8 factors - p^7 or p^3 q or p q r where p,q,r are prime factors.