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.)
Other paths... | Comment 6 of 17 |

Daniel, as I read your first posting, I assumed you had found an earlier solution (presumably the same one I found) but had just not stated it.  Your other (posted) solutions, however, do not seem to be solutions to the original problem, since that clearly required two three-digit numbers, and one four-digit number, though they were solutions to the restated (x,y,z) version.

As brianjn notes, we are faced with the precision problem when we find that there are no solutions with integer cube roots. After finding no integer roots, I tried with six-decimal precision, and got a half dozen or so that were "close" but only one which was accurate to that precision (192, 375, 2187).  The best tool at hand was the MS Calculator, which gave agreement to 18 places.  Since this was not absolute proof, I wondered if anyone found a "pencil and paper" approach which derived that set uniquely.

The cube roots I found with the Calculator were:

192 -- 5.768 998 281 229 633 529

375 -- 7.211 247 851 537 041 911

2187 -- 12.980 246 132 766 675 440

It seems that there should be a clever approach which would have multiplied the original by some common factor, and then reduced the equation to a proof.  All three of my answer numbers had a common factor of 3, removing which left respectively, 2**6, 5**3, and 3**6 but I don't see how that leads to a solution. -- or perhaps it does...

If I divide the three answers by 3, that gives 64,125,729. Taking the cube root of each of these gives 4, 5, and 9; and of course 4+5=9. Perhaps the way to build a solution, would be to start with the lowest x,y,z and then multiply that triad by a common factor.  On consideration, perhaps this was daniel's method all along??!!


  Posted by ed bottemiller on 2009-07-30 11:21:34
Please log in:
Remember me:
Sign up! | Forgot password

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

Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information