Great explanation. Yes, that is the correct solution. Congratulations!

Side note: However, this puzzle only works if the daughters love and hate the right flowers. With some constellations it would be unsolvable in practice.

OK, here is the next one. Again, initially it seems like there are too many unknown variables:

Two brothers, who were camel merchants, sold their herd of camels. Coincidentally, the price of each camel in dollars was equal to the number of camels in the herd they sold.

They received their payment in 10$ bills plus some loose change (the loose change was smaller than 10$ in total). They decided to split the money by having each one of them take a 10$ bill in turn. They did this till there was only the loose change left.

"Listen," said the older one. "I got one more 10$ bill than you did. So you take the remaining loose change. And then you take the bag in which we got the money, so that we will be completely even."

They younger brother agreed that this was fair.

How much was the bag worth?

(HINT: only find out what you're asked to.)

Seems like the only number it works with is $4 for the bag.

Correct. Your turn.

(if you want to, give more details and explanation of your solution...)

Did it rather the grunt way. Problem says number of dollars per camel is equal to number of camels, so the total must be a square integer. Then since one brother got $10 more, the tens digit of said total must be uneven. So started going through squares of numbers up to mid-20s and noticed that the only ones fitting that are squares of numbers ending in 4 or 6, both of which end in 6. Then glanced at an actual list and saw the pattern holding (actually didn't know those patterns among square numbers, can be handy).

But no idea of math puzzles of my own. (Hm, should I not try to solve them then?)

There is actually a proof for the pattern. I'll post something more on this when I have time later.

EDIT:
OK, I have some time now so here is a bit more detailed explanation.

As noted, the amount of money is a perfect square, with its tens digit being odd.

Method 1:

Some properties of perfect squares:

All perfect squares end in 1, 4, 5, 6, 9 or 00.
If it ends in 6, ten’s digit is always odd (1, 3, 5, 7, and 9), otherwise, if it ends in 1, 4, 5, or 9 the ten’s digit is always even (2, 4, 6, 8, 0).

That is enough to solve the problem. For those interested, there are more properties. You can look for proof and more properties on the web. It would be too cluttered to post it here. From Wikipedia:

If the last digit of a number is 0, its square ends in an even number of 0s (so at least 00) and the digits preceding the ending 0s must also form a square.
If the last digit of a number is 1 or 9, its square ends in 1 and the number formed by its preceding digits must be divisible by four.
If the last digit of a number is 2 or 8, its square ends in 4 and the preceding digit must be even.
If the last digit of a number is 3 or 7, its square ends in 9 and the number formed by its preceding digits must be divisible by four.
If the last digit of a number is 4 or 6, its square ends in 6 and the preceding digit must be odd.
If the last digit of a number is 5, its square ends in 25 and the preceding digits must form a pronic number.

Method 2:

With a "proof" of mine. It might not be too neat or original, there are probably better ones out there. But here goes.

Hypothesis: perfect square mod 20 gives one of the following: 0, 1, 4, 5, 9 or 16

"Proof":

Any natural number can be expressed in the following form:
10a + b
where a is a non-negative integer and b is a non-negative integer smaller than 10

e.g.
28 = 2 x 10 + 8
927 = 92 x 10 + 7
5 = 0 x 10 + 5
... etc

If we square it
(10a + b) ^ 2 = 100a^2 + 20ab + b^2

100a^2 is divisible by 20
20ab is divisible by 20

So we're left with only b^2 where b is in range 0-9
The possible values of b^2 are: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81
All these values mod 20 give one of the following: 0, 1, 4, 5, 9 or 16.

Hence proven.

Of these values, we have to disregard the ones smaller than 10, since they were left with an odd 10$ bill and loose change. Therefore this had to be 16. The elder brother took the extra 10$ bill, while the younger one took 6$ and the bag, which had to be worth 4$.

Doesn't have to be strictly maths. See if you can come up with anything. If you still have nothing, pass your turn to whoever wants to take it.

Passing it, then.

Hm... Did I "kill" this by not adding a puzzle?

I'd add something, but I'm so miserable at math that you guys would figure it out in a heartbeat.

How about this one:

A 9 Volt battery is connected to a superconducting wire (0 resistance) which is turn connected to a LED. The circuit is complete (that is, it is not broken).

How much current flows through the LED? (Hint: Use Ohm's Law.)

Again, it doesn't have to be strictly maths. Your non-maths puzzles are good, just as long as it's pretty short; not something that requires its own thread like the jury one.

dtgreene's puzzle is up.

Well, I mean, it IS a "maths puzzle" thread.

Hmmm... given the interest this got when it started I wasn't expecting it to die so quickly. So... bump?

That's what is commonly known as a short-circuit. As Ohm's Law clearly states, Intensity equals to Voltage divided by Resistance, if a Resistance is Zero, theorically Intensity should tend to infinite but as this isn't possible, in real world terms it necesarilly needs to be superior to Zero as well.

I am so rusty in maths but I will try to make a simple puzzle for you to solve:

I have a total of 48 socks in my drawer. 1/4 are red, 1/3 blue and the rest white. I am looking for the least movements needed to guarantee obtaining a matching pair if I pull out just one sock at a time.

REDVWIN

Least movements:
1) Activate sufficient light source.
2) Pull a sock.
3) Pull a matching sock.

Assuming you can't check for matches before you've finished pulling socks:

48/4 = 12 red socks
48/3 = 16 blue socks
48 - (12 + 16) = 48 - 28 = 20 white socks

To guarantee a matching pair of socks, you would need to pull 4 socks. 3 pulls could result in 3 different colors of sock, and the 4th pull would have to match one of those initial 3.


I don't have a puzzle at the moment. Someone else can post one.