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Wednesday, May 25, 2005

Part two, the hardest word problem you have ever seen in your life.

After you can do the previous problem, and find the algorithm to sort the nine coins, try this one.
This one is really gross. I only ever knew one person to get this one, not counting myself.
I would go so far as to say that this problem is harder than for a married man to balance a checkbook.
Anyway, have fun with it. Here it is:

The man in the previous problem has eleven real coins instead of eight. Add in the counterfeit for a total of twelve.
In a total of three measurements, find

1)which coin in counterfeit
2)whether the counterfeit coin is heavier or lighter than the real ones.

when you dont know either to start with.

That's it. Sound easy?
The first person who emails the ENTIRE correct algorithm will get a dollar by paypal so they can go buy themself a mars bar.

Thursday, May 12, 2005

The Ultimate Word Problem. Part One.

I know of a word problem which is, in my opinion, The Ultimate.
In order to tell it, I have to give an easier one first, just so the other will make sense.
Here's how it goes:

Suppose you had nine coins. Eight of them gold and one counterfeit. The eight gold coins are the same. The counterfeit one is almost perfect. The only way to tell the difference is by weight.
Now suppose you had a balance. The kind with a tray on either side such that if you place something on each side, the heavier side will go down and the lighter side will go up. Suppose the tray is large enough to hold as many coins as you need to place in them.
Finding the counterfeit coin becomes an easy task. Simply begin by placing a coin in each tray and if they are the same, the two trays will remain level. Then substitute another coin in one of the trays until you find which one is different.
But suppose you wanted to do it in the fewest measurements possible. It can be done in three measurements.
So the problem is: How can you determine in three measurements 1)which coin is counterfeit 2) whether the counterfeit is heavier or lighter than the real ones

Hint: Start by placing three coins in each tray.

Coming next....part two, the hardest word problem you have ever seen in your life

Monday, May 09, 2005

Another kind of problem with numbers

That first number problem needed some simple logic to figure out. The trick was that you had to go beyond the basic logic that you thought you knew. The second one used more of a problem-solving type of logic, because you need to figure out why you can't shave off enough time and then solve it accordingly. I know a lot of different ones, but a lot of them use the same type of thinking as the first two, so it would be somewhat redundant. If anyone is able to get the answer to that second one, email me and I will post your name.
I have two more old-time word problems in mind right now. One is easy, but the other one is like stupid-hard. They each use different kinds of thinking, but to me they seem to all be related.
Anyway, I'm working on the drafts for them to put up here.

Here is the first one, it uses what I would call "systematic thinking":

Suppose you had a dollar.
Now suppose you had someone make change for it.
They could give you a silver dollar. That would be one coin.
They could give you two half-dollars. Two coins.
They could give you a half-dollar and two quarters. Three coins.
Question: what is the smallest number of coins that could not add to exactly one dollar?

See, not hard. Just stubborn. Good luck.

Saturday, May 07, 2005

A harder word problem

Here is another classic word problem. It's a bit harder than the first one. Don't expect to figure it out on the first try. Or the second try. If you work by trial and error, you could, theoretically, get the answer, but I haven't ever seen anyone actually get it that way. The easiest way to figure it out is to try it a couple of times and then stop and think about it and keep asking what went wrong.
Ok, here it goes:

Four explorers are out hiking a trail at night. They come across a deep ravine with a rickety old rope bridge with half rotten wooden slats. They only have one flashlight among them. Because it is so dark and the bridge is so old, only two of them can get across safely at a time. Also, because of varying degrees of agility and bravery, they can each cross at different speeds.
The first guy can get across the bridge in one minute.
The second guy can cross it in two minutes.
The third guy in five minutes.
And the fourth guy in ten minutes.
If two people are crossing together, they have to travel at the slower person's pace.
So, for example, if they all want to cross the bridge, they could do it as follows:

Person number one and two go across together. (takes two minutes)
guy number one comes back across with the light (takes one more minute. three total so far)
guy number one and three go across. (takes five more minutes, total of eight)
number one goes back across with the light (add another minute, total of nine.)
number one and number four cross in ten minutes (grand total of nineteen minutes)

So here is the question...........................How can they all cross in only seventeen minutes?

There are actually two solutions for it. Good luck.

Friday, May 06, 2005

I like numbers after all

I remember when I was in highschool. One of the most dreadful of all test questions was always the word problem, especially in math. Do you remember those? The ones that always started off with two trains travelling toward each other at high speed, which never really made sense to me at all, since if it were true, they would both hit the brakes sooner or later.
But somehow, now that I am way old, I really kind of like them. Or at least the good ones.
There is a difference between the good ones and the bad ones. The bad ones have to do with counting the apples or figuring out how much older John is. But the good ones are different.
The good ones make you think along lines that you didn't consider at first. I'll give you an example. This one is The Classic.

Imagine you were in a stadium full of people and had everyone who had two children come foreward. Then you asked those people to raise their hand if they had at least one boy.
At that point, everyone who didn't raise their hand (those who had two girls) could go sit back down. Now out of all of those people remaining (those with two children, at least one of which was a boy) what percentage of them should have two boys?

Did you say fifty percent? If you did, you are among the oogles of people who missed that one.
I know what your logic is. You reason that if one is a boy, the other is either a boy or a girl, and since you have a fifty-fifty chance, the odds are fifty percent. Makes perfect sense. However, it is wrong.

No, it is not a trick question. The real answer is 33 1/3%.
But how is that possible?
Take another look at the question and see if you can figure it. I promise you, it's true.

The key is in how it is set up. Here's how it works:

If you got married and had two children, there are several possibilities.
1. you could have two boys.
2. you could have a boy and a girl.
3. you could have a girl and a boy.
4. you could have two girls.
So if you take away those with two girls, you have three choices remaining, with even odds for each, so one third of them should have two boys.
Now I guess you could throw in fraternal twins and skewer the odds somehow, but that's not the purpose here.

What's good about that one is that you have to stop and think about it more deeply than it appears at first.
That's one of the really easy ones, but it's quick to write, so that's why I picked it.

I'll see if I can put down a harder one next time. But usually, you are not supposed to tell the answer. Ever. The really hard ones can take days to figure out, so be warned.

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