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

Home > Numbers
Powered s.o.d. (Posted on 2015-05-20) Difficulty: 3 of 5
List all three-digit numbers such that equal the sum of their first digit and the square of their second digit and the cube of their third digit:
e.g. 135 = 1^1 + 3^2 + 5^3.

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Paper and pencil solution (spoiler) Comment 2 of 2 |
This is clearly a simple computer program, or Excel exercise.  But I felt like giving it an analytic whirl.

Let the number be abc.
Then 100a + 10b + c = a + b^2 + c^3.
Then 99a + b(10-b) + (c-c^3) = 0

Consider that equation mod 3 and mod 11.  This eliminates a as a factor.  

I won't bore you with the analysis.

Results are:
Mod 11, bc can only be 00,01,18, 24,29,35,43,63,75,84,89 or 98.

Mod 3, b can never be 2, 5 or 8.  That leaves 8 possible values of bc: 00,01,18,35,43,63,75,98.

Calculating b^2 + c^3 determines (for each) the only possible value for a.  It is interesting to me that they all lead to solutions, if we allow leading zeroes.

solutions are 000, 001, 518, 135, 043, 063, 175, and 598.

Well, that worked out better than I expected.




  Posted by Steve Herman on 2015-05-21 07:49:32
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 (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