
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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December 18th, 2016, 06:43 AM  #11 
Senior Member Joined: Apr 2008 Posts: 194 Thanks: 3 
Thanks again for your excellent solution, romsek.

January 9th, 2017, 10:15 PM  #12 
Member Joined: Oct 2016 From: labenon Posts: 33 Thanks: 4 
Here's the number theory approach! You must know that there are 25 primes less than 100 and they are: 2  3  5  7  11  13  17  19  23  29  31  37  41  43  47  53  59  61  67  71  73  79  83  89  97 (25 primes) For a number N to have exactly four factors (divisors), the number should be of the form N = p³ or pq where p and q are distinct prime numbers. Case 1: N =p³ form Clearly, p can acquire only 2 values: 2 and 3 So, there are a total of 2 numbers of the form N =p³. Case 2: N =pq form such that q>p When p=2, 0<q<50 So, there are 14 values that q can possibly have. When p=3, 0<q<33.33 So, there could be possibly 9 values for q. When p=5, 0<q<20 So, there are only 5 values of q. When p=7, 0<q<14.28 So, there are 2 values of q. We should not move ahead! Therefore, there are a total of 14 + 9 + 5 + 2 = 30 numbers of the form N =pq Thus, there are 2 + 30 = 32 numbers that have exactly 4 factors (divisors). I hope it helps! Last edited by skipjack; January 10th, 2017 at 10:02 AM. 
January 21st, 2017, 05:12 PM  #13  
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 390 Thanks: 70 
The problem is: Quote:
Quote:
 

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100, factors, find, integers, positive, positives 
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