All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
Increase in number of factors. (Posted on 2010-10-05) Difficulty: 2 of 5
If n has 15 factors (1 and n inclusive ) and 2n has 20 factors. What is the number of factors of 4n?

Also, what can be the number of factors of 5n ?

See The Solution Submitted by Vishal Gupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Possible solution | Comment 1 of 5

General

The rules for computing the number of divisors of a number are:

1 The number 1 counts as 1 divisor.    
2 For each factor, we pick the number of distinct ways it can be multiplied, e.g for 2^3 we have 2, 2*2,2*2*2 etc. Each way counts as 1 divisor, and below as 1 'choice', grouped into a 'factor group'.    
3 For each pair and greater number of  choices combined from distinct factor groups, we count 1 divisor.   
     
n and 2n
     
2 is necessarily a factor of 2n.
    
We also want n to have 15 divisors, i.e. 14 other than 1.    
Clearly, every time we add a new factor, we increase the number of factor groups, so that doubles the number of combinations.    
     
It follows from this and the general rules that n cannot have more than 2 distinct factors, and if 2 is not a factor of n then it will double the number of divisors when it becomes a factor of 2n.   
     
Also, 2 cannot be the sole factor of n since increasing the power of 2 only increases the number of divisors by 1, if there is no other factor. 
     
Call the other prime factor P:    
 - for 2*P we have 4 divisors    
 - for 2*P^2 we have 6 divisors    
 - and for 2*P^x we have 2x+2 divisors    
When the power of 2 is increased, we add a choice to the combination between factor groups i.e. 1 for the choice of the new multiple of 2 and x for the different choices of P.    
     
     
It follows that we are seeking a 4th power of P (so that 5 new divisors will be added when n is doubled). The smallest n that meets these requirements is 2^2*3^4, or 324; but any P will work just as well.

4n
    
If we increase the number to 4n, that is another multiple of 2, so we are adding another choice and hence another 5 divisors, to a total of 25. 

5n 

If we increase the number to 5n, then either P was 5 or it was not 5.    

If P was 5, then we increase the power of 5 by 1, so the number of divisors is increased by 3, one for the new choice, and one for each of the two ways it can combine with the {2,4} factor group. There will be 18 divisors.   
     
If P was not 5, then we are adding a new factor group, so the number of divisors is doubled, to 30. 

 

Edited on October 5, 2010, 2:58 pm
  Posted by broll on 2010-10-05 14:27:58

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (14)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information