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

Home > Just Math
Testy Triplet Treat (Posted on 2013-08-04) Difficulty: 3 of 5
P(N) denotes the product of the digits in the base ten representation of N.

Determine all possible triplets (A,B,C) of positive integers with A ≤ B ≤ C that satisfy this system of equations:

A = P(B)+ P(C), B = P(A) + P(C), and C = P(A) + P(B)

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 re(2): results so far, but no proof in sight | Comment 3 of 7 |
(In reply to re: results so far, but no proof in sight by Charlie)

My results were : 5,5,10, found while thinking of a zero digit and

36,36,36  based a A,A,A assumption and solving for a 2-digit case.
1OP+Q=2*P*Q  ==> Q=10P/(2P-1 )  ==>P=3  Q=6.

I was almost sure that no other examples exist but decided
to wait  for other solvers.

I still do not see how  one can prove that  there are no other 

  Posted by Ady TZIDON on 2013-08-04 15:25:36
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 (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

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