Guest Posted March 9, 2010 Report Share Posted March 9, 2010 (edited) Determine all possible quintuplet(s) (A, B, C, D, E) of positive integers, with A <= B <= C <= D <= E, that satisfy this equation: A*B*C*D*E = (A+1)+(B+2)+(C+3)+(D+4)+(E+5) Edited March 9, 2010 by K Sengupta Quote Link to comment Share on other sites More sharing options...

0 plainglazed Posted March 9, 2010 Report Share Posted March 9, 2010 (edited) EDIT: it's late, better rethink this... Edited March 9, 2010 by plainglazed Quote Link to comment Share on other sites More sharing options...

0 plainglazed Posted March 10, 2010 Report Share Posted March 10, 2010 Okay, time for another stab - notice the five variables cannot be all even or all odd - therefore, the left hand product will always be even - consequently the five constants must contain exactly one or three odd numbers from here my solution trades elegance for brute force: First with three odd numbers among the five: consider (1,1,1,D,E) ==> DE=18+D+E or E(D-1)=18+D - for D=2, E=20 so (1,1,1,2,20) is a solution - for all other even D resulting in E>2, E is not an integer consider (1,1,3,D,E) ==> 3DE=20+D+E or E(3D-1)=20+D - for D=2, 5E=22, E is not an integer - for D=4, 11E=24, E is not an integer - for D=6, 17E=26, E is not an integer and E<2 consider (1,1,5,D,E) ==> 5DE=22+D+E or E(5D-1)=22+D - for D=2, 9E=24, E is not an integer - for D=4, 19E=26, E is not an integer and E<2 consider (1,1,7,D,E) ==> 7DE=24+D+E or E(7D-1)=24+D - for D=2, E=2 so (1,1,2,2,7) is a solutin and with any D>2, E<2 consider (1,3,3,D,E) ==> 9DE=22+D+E or E(9D-1)=22+D - for D=2, 17E=24, E is not an integer and E<2 consider (1,3,5,D,E) ==> 15DE=24+D+E or E(15D-1)=24+D - for D=2, 29E=26, E is not an integer and E<2 consider (3,3,3,D,E) ==> 27DE=24+D+E or E(26D-1)=24+D - for D=2, 51E=26, E is not an integer and E<2 and all other A,B,C,D will also result in E<2 Now with only one odd number among the five variables: consider (1,2,2,D,E) ==> 4DE=20+D+E or E(4D-1)=20+D - for D=2, 7E=22, E is not an integer - for D=4, 15E=24, E is not an integer and E<2 consider (3,2,2,D,E) ==> 12DE=22+D+E or E(11D-1)=22+D - for D=2, 21E=24, E is not an integer and E<2 consider (1,2,4,D,E) ==> 8DE=22+D+E or E(7D-1)=22+D - for D=4, 31E=26, E is not an integer and E<2 and all other A,B,C,D will also result in E<2 so am thinking the possible (A,B,C,D,E) are (1,1,1,2,20) and (1,1,2,2,7) Someone please simplify this. Quote Link to comment Share on other sites More sharing options...

0 Guest Posted March 11, 2010 Report Share Posted March 11, 2010 For the past hour I was expanding, simplifying, dividing, multiplying and all kind of stuff with x^{2} + y^{2} + 1 = x*y*z. I somehow found that xy^{-1} - 3 = z^{2}. From here: z^{2} must be greater or equal to 0 (To be a real number). From the original equation we find, that z must be greater or equal to 3, so for any z greater or equal to 3 there is a xy^{-1} - 3, equal to z^{2}. Quote Link to comment Share on other sites More sharing options...

0 Guest Posted March 12, 2010 Report Share Posted March 12, 2010 Sorry, I posted to the wrong topic. Ignore the upper post. Quote Link to comment Share on other sites More sharing options...

## Question

## Guest

Determine all possible quintuplet(s) (A, B, C, D, E) of positive integers, with A <= B <= C <= D <= E, that satisfy this equation:

A*B*C*D*E = (A+1)+(B+2)+(C+3)+(D+4)+(E+5)

Edited by K Sengupta## Link to comment

## Share on other sites

## 4 answers to this question

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.