Riddles

[Herremann the Wise]

The following is one of the most beautiful puzzles I have ever seen. It is neither complex nor too easy and can be asked to any school kid who has their head around multiplication up to as old as you care. (8 to 88 if you like).

The solution is beautiful in its pure elegance.

The following is a version of it I made into rhyming stanzas for game purposes.
Good luck and kudos to the first person who can solve it.

There were once two friends of old
Who met again after many a year
One greeted the other warmly
And of three daughters did he hear.

However, riddles he did like
And so a riddle he was given
To work out the Daughters ages
And thus what he’d been missing.

His friend was quick of wit
And so the riddle hard
Of only three clues would he give
And so keep his friend on guard.

“The first clue is simple
And shall contain no feisty tricks
For their ages when multiplied
Amount to Thirty Six.”

To this his friend thought
And smiled and nodded freely
This puzzle he was meant to solve
Would really be too easy

“Now the second clue relies
A little on your memory
Of when we were young
And paid silver pieces for a pony.”

“If you can remember,
How much we paid for ‘Feather’
It is the same as the total when
You add their years together.”

Of this Pony’s price he could remember,
But confused was he too
For really this did not help
He’d need another clue.

And with a smile his friend said
“A final clue will I give thee
I’m told that my Eldest Daughter
Looks just the same as Me”

Simple, just tell me the ages of the three daughters.

Best Regards
Herremann the Wise

 

[RangerWickett]

Simple, just tell me the ages of the three daughters.

Best Regards
Herremann the Wise

First, your rhyme scheme and meter could use some serious work. It just doesn't flow off the tongue as is, unless I'm missing some sort of strange beat.

Second, their ages are 2, 2, and 9.

Third, the explanation is actually rather simple. We have three clues:

Clue one is that the product of the three daughters' ages is 36.

Clue two is that if you add their ages together, it provides an ambiguous answer, since that clue alone didn't help him. What this means is that there have to be at least two sets of possible ages that have the same sum. Let's assume for simplicity's sake that the person isn't using fractional ages for his children, because if he is, he deserves to be punched.

So we have options like:
1, 1, 36 = 38
1, 2, 18 = 21
1, 3, 12 = 16
1, 4, 9 = 14
1, 6, 6 = 13
2, 2, 9 = 13
2, 3, 6 = 11
3, 3, 4 = 10

There are only two sets that have the same sum. 1, 6, 6 and 2, 2, 9, both of which have a sum of 13.

Now comes the third clue. It doesn't matter that a daughter looks like him. What matters is that he has an "eldest" daughter, meaning that the twins can't be the older set. If they were 1, 6, and 6, then he'd have no eldest daughter. Okay, he would have an eldest daughter, but the difference would be a matter of minutes. So we know that the twins must be younger, which gives us the set of 2, 2, and 9.

 

[der_kluge]

I found a riddle, or puzzle rather, a while back, but never found the answer to it. Maybe someone here knows it.

It was something like this:

You come into a room with 12 balls. Each ball is identical in size and shape. One of the balls has a different weight than the all the other balls.

Using a scale in the room, can you find the one ball whose weight does not match the others if you can only use the scale 3 times?

the ball can weigh either more or less than the other balls, you do not know which it is.

Sorry - I don't actually have the answer to this one, and it has completely stumped me, so if someone does know it, I'd be much appreciated.

 

[333 Dave]

Oops. See Below.

 

[333 Dave]

I found a riddle, or puzzle rather, a while back, but never found the answer to it. Maybe someone here knows it.

It was something like this:

You come into a room with 12 balls. Each ball is identical in size and shape. One of the balls has a different weight than the all the other balls.

Using a scale in the room, can you find the one ball whose weight does not match the others if you can only use the scale 3 times?

the ball can weigh either more or less than the other balls, you do not know which it is.

Sorry - I don't actually have the answer to this one, and it has completely stumped me, so if someone does know it, I'd be much appreciated.

I did this with 8 pills and 2 uses of the scale (way back in 4th or 5th grade). Lemme crack at 12 and 3.

Set two aside. This will be Group A.
Divide the remaining 10 into Groups B and C, 5 each.
Balance B and C.
If B > C,                            //Balance use 1
        Divide B up into two groups of 2 each and one spare.
        Balance B1 and B2.   //Balance use 2
        If B1> B2               
             Divide B1 into two groups of one each.
             Balance B1a and B1b.
             The heavier is the odd one
        If B2 > B1, repeat above steps
        If B1 = B2, B3 (the spare when dividing up B) is the odd one.
If C > B, do steps for B with C instead.
If B == C,
       Balance Group A's two members. The heavier one is the odd one.

 

[333 Dave]

On a side note, thank you die_kluge, for posting a question that could make me feel smart. I really needed that today.

 

[333 Dave]

OH! Didn't see that notion that the ball could be heavier OR lighter! That requires extra steps! Gimme a minute....

 

[Herremann the Wise]

You have 12 balls, one of which is heavier.

Split into two groups of 6. One will be heavier and this will have the weighted ball so you then:
Split that group of six into two groups of three. Again you take the set of balls in it that's heavier.
You are now left with three balls one of which is weighted and you have one use of the scales left.
Pick any two out of the three balls and put them on the scales. If one of them is heavier, you know that's the weighted ball. If the scales balance, then you know it must be the ball that was left out.

[Edit: Ha. Ignore the above, I fell for the same thing! )

Best Regards
Herremann the Wise

PS:
Good stuff RangerWickett. Unfortunately, most people miss out on the hidden fourth clue - that the sum is ambiguous. Its a good not too easy, not too difficult puzzle. (And apologies for the writing, my skill in prose far outweighs my skill in poetry ).

 

[333 Dave]

Ok, I think that my solution still works except for when you have to balance B1A and B1B.
For Group A to work, just add a statement where you balance one of them against another non-group A ball, if equal the other non-weghed group A ball is odd, if unequal, the one from group A you weighed twice is the odd one.
But I need another use of the balance (like I used for A) if it goes that far into B. Lemme think more on the topic...

EDIT: Ambiguity Clarified.

 

[333 Dave]

EURIKA!
Instead of Comparing B1A to B1B, Compair B1A to any other ball. If they're equal, then B1B is the odd one. If not, B1A is! HAHA! TAKE THAT WEIGHTS AND MEASURES PUZZLE!