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

Home > Just Math
Cubic Equation (Posted on 2013-06-09) Difficulty: 4 of 5
Determine all triplets (a,b,c) of nonzero integers satisfying:
987654321*a + 123456789*b + c = (a + b + c)3

Prove that there are no others.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts preliminary computer results (spoiler) | Comment 1 of 13

Let x = a + b + c

987654321*x > x^3
987654321 > x^2
x < 31427

x^3 > 987654321
x > 995


DEFDBL A-Z
FOR t = 995 TO 31427
    IF t MOD 1000 = 0 THEN PRINT t;
    FOR a = 1 TO t - 2
        FOR b = 1 TO t - a - 1
            c = t - a - b
            lhs = 987654321 * a + 123456789 * b + c
            sum = a + b + c
            rhs = sum * sum * sum
            IF lhs = rhs THEN
                PRINT a; b; c, lhs
            END IF
        NEXT
    NEXT a
NEXT t

has found so far

 a      b      c
450   4500   5050  results in 1000000000000 for each side.

It's still chugging away and has checked beyond a+b+c = 12000 so far, but not yet up to 13000.


  Posted by Charlie on 2013-06-09 15:01:35
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