Logic Puzzles

oh no shit, i see your point, i'm being a div.

I've spent most of my day today on this thread being a div.

I think I got this although i'm making the assumption that the clever man is in fact clever and that he knows the house number and so has tried to work it out with that knowledge and failed.

Applying the first clue: Numbers that multiply together to for you to get 36 are as follows:

1, 1, 36
1, 3, 12
1, 4, 9
1, 2, 18
1, 6, 6
2, 2, 9
2, 3, 6
3, 3, 4

Second clue: (The second man seems to know the door number but this clue does not help him. Therefore i'm assuming after adding up all the combinations the man found some to give the same number otherwise he'd have guessed the ages) The sum of the different combinations are as follows:

1, 1, 36 = 38
1, 3, 12 = 16
1, 4, 9 = 14
1, 2, 18 = 21
1, 6, 6 = 13
2, 2, 9 = 13
2, 3, 6 = 11
3, 3, 4 = 10

The sum of the combinations are all different except for 1, 6, 6 & 2, 2, 9 which both equal 13. So it's one of these. If not then knowing the house number the clever man would have picked it. But because it's the sum of two combinations he's still unsure.

Third clue tells us that there's a younger daughter (thanks horatio!). So it's not 2, 2, 9.

Therefore their ages are 1, 6 and 6. And the house number is 13.

...I think.

Confirmation! I call this case closed.

Now can someone explain the dog one to me. I really want to figure it out. I get most of it but not the restrictions on the dog. Is the dog on a leash that is of a certain length? or is the dog in effect on a track that closley follows the perimeter of the fence.

In other words is it like imagining that there is a thick circle drawn out on the floor in white and the dog is only allowed to stand on the white line?

Also, if the above is true, can the dog travel in both directions?

I stopped on the dog one as soon as it was clear you needed to use mathematics. Pshaw.

That's what I like! I hate the word play ones. Never get them.

it's nothing clever with the dog - it's just pure maths. imagine the dog just as a dot, right on the fence, can move in both directions at constant speed.

aaaaargh! Why did you post the answer!!!

And so unusually concise! Getting that answer into your brain box before you even know what's what!

I meant the question... hence why I was seeking clarification on the dog's path.

yes and yes.

Isn't that pretty much what I said (without the 0.25R circle bit)?

Third clue tells us that there's a younger daughter (thanks horatio!). So it's not 2, 2, 9.

Only just seen this one but have a couple of thoughts.

If the father had 2 year old twins, wouldn't you expect him to know which was born second and therefore the younger of the two?

He could also have two 2 year old daughters who weren't twins.

Here's one of the old-school logic problems that I used to do on wet holidays in Devon. Man, I was a sad kid. I've tailored it a bit to suit the forum.

*Five Biking Fools visted LFGSS and were keen to try out some of the forum rides. What were the fools' names, what was the name of the bike they rode first and how many points out of five did they give it?

(Points column must contain numbers 1,2,3,4,5)*

Jeez gave one more point to the bike that he tried first, than the person who rode the Time Machine first. Was this Snottyotter?

Skrtluv's bike, which turned out to be a bit of a shitter, was awarded 2 less points than the bike rode first by umop3pisdn, which was not the Elephant Explorer or the Oak Randonneur.

TM's Corima was tried first by a girl and scored one more point than the bike rode first by Scoot.

Shoosh was thrilled with her first ride, not the Elephant Explorer, and gave it the highest number of points.

Only just seen this one but have a couple of thoughts.

If the father had 2 year old twins, wouldn't you expect him to know which was born second and therefore the younger of the two?

He could also have two 2 year old daughters who weren't twins.

This is why I like purely maths based puzzles. Because you can assume that there are no small discrepancies like that or word play. It's all straight forward.

At least some of this must be correct:

Skurluv's bike get's a 1
Umop3pisdn rides the time machine first giving it 3
Scoot gives her first ride a 4
Shoosh rides the Corima, she gives it a 5

I think something is wrong but it's too late now. Will ry again in the morn

No.

Won't work anyway since twins are born within an hour. Usually 17 minutes or so. It also won't work if you then worked it out in decimals to the nearest minute because door numbers don' come in decimals unfortunately.

14b, the maths is all in hexadecimal.

Dredge.

Is there a solution like this Pentomino one where none of the pieces need to be rotated in mirror symmetry?

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where none of the pieces need to be rotated in mirror symmetry?

Can you explain that again?

for example, the green 'P' shaped piece in the top left of the 6x10 grid cannot be rotated into place in the 5x12 grid, instead it has to be fipped over.

So you want a set of pieces that can make up x × y rectangles without any of them needing to be flipped over?

I'm 99% sure that is possible based on the following "proof": For example you could mirror every grid vertically, so that the first set becomes a 6×20 grid. For the left hand side of the 5×24 grid you select all the pieces with the correct handedness leaving a mirror set for the right hand side.

The problem with that is you get duplicate pieces, is that something you want to avoid?

duplicates would be bad. these 12 different shapes all fit together to form 6x10, 5x12, 4x15 and 3x20 rectangles with no duplicates or gaps etc. there are multiple ways of solving each individual rectangle. what i want is a solution for all 4 rectangles where none of the tiles need to be flipped over.

to put it another way, i want to make one of the solutions, colour in the top faces and then make the other solutions without having to turn any of the pieces over - in this case i made the 3x20, drew some lines, then when I come to make the 4x15 the 'P' needs to be turned over.

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