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

Home > Numbers
N-Divisibility (Posted on 2004-02-29) Difficulty: 3 of 5
How many positive integers divide at least one of 10^40 and 20^30?

See The Solution Submitted by DJ    
Rating: 4.1111 (9 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Puzzle Solution With Explanation Comment 4 of 4 |

Let D(x) denote the total number of divisors of x, including 1 and itself.

Thus, taking m = 10^40 = (5^40)*(2^40), and:
n = 20^30 = (2^60)*(5^30), we have:

p = GCD(m, n) = (2^40)*(5^30)

Thus, the required number of divisors (v) dividing at least one of 10^40 and 20^30 is given by:

V = D(m)+ D(n) - D(p)
= 41^2 + 61*31 - 41*31
= 2301

 


  Posted by K Sengupta on 2007-05-28 11:55:32
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 (7)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information