Unutterably Silly Quiz

[PeterT]

Well one simple means would be to say

"PDurrant; please arrange the coins in two stacks for me making sure that each has the same number of heads up coins".

[Daithi]

Just to be clear, you could tell me that there are 5 coins heads-up and I will be able to place them into 2 stacks that have the same number of heads-up coins?

[issybird]

I can get partway there. If there's an odd number of heads, you have to fix that first. If it's an odd number of heads in an odd number of coins, then turn all the coins over. If it's an odd number of heads in an even number of coins, you turn over one coin.

But I haven't figured out how to stack them yet.

[pdurrant]

Originally Posted by Daithi

Just to be clear, you could tell me that there are 5 coins heads-up and I will be able to place them into 2 stacks that have the same number of heads-up coins?

Originally Posted by issybird

I can get partway there. If there's an odd number of heads, you have to fix that first. If it's an odd number of heads in an odd number of coins, then turn all the coins over. If it's an odd number of heads in an even number of coins, you turn over one coin.

But I haven't figured out how to stack them yet.

Yes, if I was told that there were five coins head-up, I would be able to stack the coins so that there were the same number of heads up in each stack.

There's no need to turn the whole stack over for an odd number of heads.

There is a simple way to do this, that works for any number of heads and any number of coins in the stack. (Well, there must be at least two coins in the stack, of course. It would be a bit difficult to make two stacks out of one coin!).

Hint: I didn't say that the two stacks had to have the same number of coins in them, just the same number of coins heads-up.

[pdurrant]

Originally Posted by PeterT

Well one simple means would be to say

"PDurrant; please arrange the coins in two stacks for me making sure that each has the same number of heads up coins".

No, that's not the answer.

And to reply to a quibble from my son when told this puzzle, as well as a blindfold, consider that you are also wearing gloves - you can't tell which are heads up and which are tails up by feel!

[Daithi]

Nice job issy. I hadn't thought of flipping the coins over.

[Daithi]

So if you have an odd number of heads up coins you need to flip one coin over.

If you flipped a heads up coin over you now have one fewer heads up coin, so an even number of heads up coins.

However, if you flipped a tails up coin over you will now have one more heads up coin, so an even number of heads up coins.

Like issy, now I'm stuck getting an equal number of heads up coins into two stacks.

[issybird]

Of course flipping one coin works whenever there's an odd number of heads!

[pdurrant]

I think it's now safe for me to hint that the solution usually involves flipping some coins.

[Daithi]

I think I have gotten a little further. I believe we are on the wrong track just flipping one coin. Instead, make a pile of coins containing the same number of heads up coins that was given. Then flip this pile of coins upside down, and both piles will now contain the same number of heads up coins. Here is an example--

Suppose you start with 5 coins with 3 heads up. Your stacks must start as one of these
3h -- 2t
2h1t -- 1h1t
1h2t -- 2h

After flipping the first stack you will have
3t -- 2t
2t1h -- 1h1t
1t2h -- 2h

It looks like it works for any number of coins as long as you know the initial number of heads up coins. I've done it for 3, 4, 5, and 6 coins with different starting heads up numbers. I'm still trying to put together a mathematical reason for why it works.

[issybird]

^^Nice work.

[Daithi]

Finally, here is the reason why it works.

Let,
H=total # of heads up coins
x=number of unknown heads up coins we grabbed

So,
heads up coins in stack A = x
heads up coins in stack B = H-x

After flipping stack A we have

heads up coins in stack A = H-x
heads up coins in stack B = H-x

I don't know how good this explanation is, but it does make sense to me. When you flip stack A you get just the opposite of the number of heads ups coins than you grabbed. Also, since stack A contained H number of coins, it can't contain too many or too few coins (this is why flipping stack B won't work -- it doesn't have the right number of coins).

Maybe pdurrant can do a better job of explaining why it works than I have done.

One example with the mathematical explanation.

Suppose you have 100 coins and 15 are heads up. You grab 15 coins at random to form stack A. There are x number of heads up coins in this stack. Let's say that you had grabbed 3 of the heads up coins. Therefore, stack A contains 3 heads up coins, and stack B contains 12 heads up coins (15-3=12). When you flip stack A, which contains 15 coins, then your 3 coins that were heads become tails, and the 12 tails become heads. So, stack A goes from x heads up coins to H-x (15-3=12).

[pdurrant]

Dathi has the right answer, and for the right reason. Next question, please Dathi!

As Dathi said, the answer is that you split the one pile into two piles, one containing the same number of coins as you've been told there are heads up coins, and the other pile containing the rest. But every coin put into the first pile must be turned over.

In the special case of being told that none of them are heads up, just split the pile into two any way you like, without turning any of the coins over.

In the special case of being told that all of them are heads up, turn all the coins over and then split into two piles any way you like.

Otherwise it works just like Dathi says. I'm don't think I could explain it better.

Originally Posted by Daithi

Finally, here is the reason why it works.

Let,
H=total # of heads up coins
x=number of unknown heads up coins we grabbed

So,
heads up coins in stack A = x
heads up coins in stack B = H-x

After flipping stack A we have

heads up coins in stack A = H-x
heads up coins in stack B = H-x

I don't know how good this explanation is, but it does make sense to me. When you flip stack A you get just the opposite of the number of heads ups coins than you grabbed. Also, since stack A contained H number of coins, it can't contain too many or too few coins (this is why flipping stack B won't work -- it doesn't have the right number of coins).

Maybe pdurrant can do a better job of explaining why it works than I have done.

One example with the mathematical explanation.

Suppose you have 100 coins and 15 are heads up. You grab 15 coins at random to form stack A. There are x number of heads up coins in this stack. Let's say that you had grabbed 3 of the heads up coins. Therefore, stack A contains 3 heads up coins, and stack B contains 12 heads up coins (15-3=12). When you flip stack A, which contains 15 coins, then your 3 coins that were heads become tails, and the 12 tails become heads. So, stack A goes from x heads up coins to H-x (15-3=12).

[Daithi]

The following quiz is especially for Issybird, and the quiz is about two actual historial people.

George and Evelyn never met but they carried on writing until late in life. It has been said that Evelyn loved George, but she was too old for him. George married in 1880. He converted to Catholicism in 1930. During World War II, he served with the Royal Marines. Partly in recognition of this, Evelyn's subsequent writings analyzed the character of that war.

Evelyn died in 1966 in Somerset. Her first full-length novel had been published in 1859. She is buried in Highgate Cemetery. He died at age 62, after having published his autobiography in 1964. He lived one year longer than she did.

How can all these statements be true?

[SneakySnake]

You didn't say they were writing to each other. hmmmmmmmm