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

Home > Numbers
How and When (Posted on 2009-07-29) Difficulty: 2 of 5
Solve this alphametic, where each of the capital letters in bold denotes a different decimal digit from 0 to 9. None of the numbers can contain any leading zero.

3√(HOW)+ 3√(AND) = 3√(WHEN)

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): two views (UBASIC) | Comment 8 of 17 |
(In reply to re: two views by Charlie)

to arrive at the solution I used a more generalized form of the problem which is finding integers x>y and z such that x^(1/3)+y^(1/3)=z^(1/3).  I used the following code

10 for x=3 to 1000

20 for y=2 to x-1

30 z=int((x^(1/3)+y^(1/3))^3)

40 if x^(1/3)+y^(1/3)=z^(1/3) and int(x^(1/3))^3<>x and int(y^(1/3))^3<>y then

50 :print x;y;z

60 next y

70 next x

this produces the sole solution x=375 y=192 z=2187

which fits as a solution to the alphametic.  This approach lead to my observation that this is acutally the first solution to the more generalized form stated above and if you change line 10 to search for values of x,y above 1000 you will find the other solutions that I previously listed.  I found this interesting because initially going into this problem I thought there would be much smaller (i.e single digit) values of x,y that gave an integer value for z.

Also the precision problems that you were incountering are most likely avoided by UBASIC because it natively uses extremely high precision arithmatic.


  Posted by Daniel on 2009-07-30 20:29:21
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