Had a bit of a time with this one, came up with many answers that were oh-so-close before finding:
5 8 3 1 7 4 6 2 9 0
No idea if there are other solutions... (Kinda doubt it, the above pairings is the only way I could keep the 8/1 pair away from the 7/2 pair...) Interesting problem! Had to devise a ball-and-stick notation that looked like this:
(click thumbnail for full-size image)
Recently yanked a 'Xirlink' camera out of an old Internet kiosk that's being retrofitted, well happy to have a new toy to experiment around with :)
I apologize for not getting another puzzle going yet. I've been busy painting walls and pulling carpet (Theresa said, "Let's refinish the hardwood flooring underneath!" ::sigh::) here at the house. I'll post one someday...
Find positive integers a, b, c, d such that a! * b! = c! * d!
To avoid trivial and symmetric cases assume c+1 b and d>a)
Also note that if c=1 then this reduces to Part 1.
There are only 4 solutions for d <= 170; it is not known by the author of this puzzle if there are any larger ones.
Find positive integers a, b, c, d such that a! * b! = c! * d!
To avoid trivial and symmetric cases assume c+1 b and d>a)
Also note that if c=1 then this reduces to Part 1.
There are only 4 solutions for d <= 170; it is not known by the author of this puzzle if there are any larger ones.
Well, this is the first one so far that has me stumped. I started by taking the log of both sides hoping to be able to work with addition instead of multiplication. Shortly thereafter I started looking around the Internet for a better (or proper) way to go about solving it, and it didn't take long to pull up the actual author's solutions. Oops. So I'll disqualify myself on this one... still not exactly sure how to go about it... from this, it looks like you have to make some assumptions on the possible intervals and then divide by the smaller factorial... That was a good one nevermorestr!
Find positive integers a, b, c, d such that a! * b! = c! * d!
To avoid trivial and symmetric cases assume c+1 b and d>a)
Also note that if c=1 then this reduces to Part 1.
There are only 4 solutions for d <= 170; it is not known by the author of this puzzle if there are any larger ones.
Well, this is the first one so far that has me stumped. I started by taking the log of both sides hoping to be able to work with addition instead of multiplication. Shortly thereafter I started looking around the Internet for a better (or proper) way to go about solving it, and it didn't take long to pull up the actual author's solutions. Oops. So I'll disqualify myself on this one... still not exactly sure how to go about it... from this, it looks like you have to make some assumptions on the possible intervals and then divide by the smaller factorial... That was a good one nevermorestr!
Admittedly I could never solve most of these or it would take me quite some time. Math was never one of my strong suits; I made it through algebra and geometry but most of it didn't stick with me (esp. all those theorems and postulates).
This is a tough problem I can find lots of solutions where B =(D-1). The solution I suspect lies in a algebraic solution to 'range in' the individual factorials based on the notes provided. My Math is to rusty. I am having a hard time deciphering the notes I can't even get to first base on this one.
There are some who can live without wild things and some who cannot. - Aldo Leopold
This is a tough problem I can find lots of solutions where B =(D-1). The solution I suspect lies in a algebraic solution to 'range in' the individual factorials based on the notes provided. My Math is to rusty. I am having a hard time deciphering the notes I can't even get to first base on this one.
Another in the vein of the first puzzle in this thread (Einstein's):
The four people in this puzzle all competed in different classes of dog agility at a recent competition. The competitions all required the dogs to run over jumps, through tunnels and various other obstacles in as quicker time as possible. Each had a different result - one came first, one third, one fourth and one ninth. All four dogs were each of a different breed.
1. If Tiff finished first then Terry finished fourth.
2. If Terry finished fourth then Jago is a collie otherwise Jago is not a collie.
3. If Jane competed in the Senior class then she finished third.
4. If Jane competed in Novice then she finished fourth.
5. The dog that finished ninth was an alsatian. This was either Jago, in which case Jago competed in the Elementary class, or this was Kelly, in which case Terry handled Kelly.
6. Mark won Starters.
7. If Mark's dog is called Patti then Patti is a labrador otherwise Patti is a collie.
8. Ruth's dog is called Jago.
9. If Jago finished fourth then she competed in the Novice class otherwise she competed in the Senior class.
10. If Patti finished first then Terry's dog is an alsatian otherwise Terry's dog is a collie.
11. If Jane's dog is a doberman then Jane finished fourth otherwise Jane finished third.
Handler's Names: Jane, Mark, Ruth and Terry
Dog's Names: Tiff, Patti, Jago and Kelly
Breed: Alsatian, Collie, Labrador and Doberman
Level: Starters, Elementary, Novice or Senior
Can you work out who handled which dog, at what level each competed, the place each finished in and the breed of each dog?
Had a bit of a time with this
)
Had a bit of a time with this one, came up with many answers that were oh-so-close before finding:
5 8 3 1 7 4 6 2 9 0
No idea if there are other solutions... (Kinda doubt it, the above pairings is the only way I could keep the 8/1 pair away from the 7/2 pair...) Interesting problem! Had to devise a ball-and-stick notation that looked like this:
(click thumbnail for full-size image)
Recently yanked a 'Xirlink' camera out of an old Internet kiosk that's being retrofitted, well happy to have a new toy to experiment around with :)
RE: This one is killing
)
There are some who can live without wild things and some who cannot. - Aldo Leopold
Good work Chipper Q and Rod
)
Good work Chipper Q and Rod Gallant!
I'll scrounge around for another problem later today.
I apologize for not getting
)
I apologize for not getting another puzzle going yet. I've been busy painting walls and pulling carpet (Theresa said, "Let's refinish the hardwood flooring underneath!" ::sigh::) here at the house. I'll post one someday...
Find positive integers a, b,
)
Find positive integers a, b, c, d such that a! * b! = c! * d!
To avoid trivial and symmetric cases assume c+1 b and d>a)
Also note that if c=1 then this reduces to Part 1.
There are only 4 solutions for d <= 170; it is not known by the author of this puzzle if there are any larger ones.
RE: Find positive integers
)
Well, this is the first one so far that has me stumped. I started by taking the log of both sides hoping to be able to work with addition instead of multiplication. Shortly thereafter I started looking around the Internet for a better (or proper) way to go about solving it, and it didn't take long to pull up the actual author's solutions. Oops. So I'll disqualify myself on this one... still not exactly sure how to go about it... from this, it looks like you have to make some assumptions on the possible intervals and then divide by the smaller factorial... That was a good one nevermorestr!
RE: RE: Find positive
)
Admittedly I could never solve most of these or it would take me quite some time. Math was never one of my strong suits; I made it through algebra and geometry but most of it didn't stick with me (esp. all those theorems and postulates).
This is a tough problem I
)
This is a tough problem I can find lots of solutions where B =(D-1). The solution I suspect lies in a algebraic solution to 'range in' the individual factorials based on the notes provided. My Math is to rusty. I am having a hard time deciphering the notes I can't even get to first base on this one.
There are some who can live without wild things and some who cannot. - Aldo Leopold
RE: This is a tough
)
Thanks for giving it a try Rod and Chip.
Another in the vein of the
)
Another in the vein of the first puzzle in this thread (Einstein's):
The four people in this puzzle all competed in different classes of dog agility at a recent competition. The competitions all required the dogs to run over jumps, through tunnels and various other obstacles in as quicker time as possible. Each had a different result - one came first, one third, one fourth and one ninth. All four dogs were each of a different breed.
1. If Tiff finished first then Terry finished fourth.
2. If Terry finished fourth then Jago is a collie otherwise Jago is not a collie.
3. If Jane competed in the Senior class then she finished third.
4. If Jane competed in Novice then she finished fourth.
5. The dog that finished ninth was an alsatian. This was either Jago, in which case Jago competed in the Elementary class, or this was Kelly, in which case Terry handled Kelly.
6. Mark won Starters.
7. If Mark's dog is called Patti then Patti is a labrador otherwise Patti is a collie.
8. Ruth's dog is called Jago.
9. If Jago finished fourth then she competed in the Novice class otherwise she competed in the Senior class.
10. If Patti finished first then Terry's dog is an alsatian otherwise Terry's dog is a collie.
11. If Jane's dog is a doberman then Jane finished fourth otherwise Jane finished third.
Handler's Names: Jane, Mark, Ruth and Terry
Dog's Names: Tiff, Patti, Jago and Kelly
Breed: Alsatian, Collie, Labrador and Doberman
Level: Starters, Elementary, Novice or Senior
Can you work out who handled which dog, at what level each competed, the place each finished in and the breed of each dog?