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

Home > Numbers
Age Difference (Posted on 2003-03-02) Difficulty: 3 of 5
Said Albert to Bertrand, "The last time we met, our ages were both prime numbers, and when I was a quarter of the age I am now, you were that age plus half the age your father would have been thirty years previous to when he was six times the age you would have been when I was half your age".

How old were Albert and Bertrand the last time they met?

See The Solution Submitted by Ravi Raja    
Rating: 3.5556 (9 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 6
Let x be Bs age when he was twice As age (when A was half Bs age).

Then the age difference when A was 1/4 his present age (and therefore the constant age difference as that does not change) is (6x-30)/2, or 3x-15. But since difference held also when B was twice As age, and B was x years old then, it follows that x=2(3x-15); that is, if B was twice as old as A, then his age was also twice the difference in their ages.

Solving this for x, we get x=6, so the age difference is 3.

In order for two prime numbers to differ by 3, they must be 2 and 5, as 2 is the only even prime.

So the last time they met they were 2 and 5.

  Posted by Charlie on 2003-03-02 10:15:14
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 (3)
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