BrainDen.com - Brain Teasers
• 0 ## Question 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

## Recommended Posts

• 0

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)

##### Share on other sites

• 0 For the past hour I was expanding, simplifying, dividing, multiplying and all kind of stuff with x2 + y2 + 1 = x*y*z. I somehow found that xy-1 - 3 = z2. From here:

z2 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 z2.

##### Share on other sites

• 0 Sorry, I posted to the wrong topic. Ignore the upper post.

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account. ×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.