Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
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.
It's just my thought of how to look at this problem from another angle but it's not really advisable in real test
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)
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.