a card problem

admin — Sun, 26 Sep 2010 05:33:15 +0000

Here is a problem quoted from fakalin. A full deck of cards has 52 cards. Suppose 10 of them were face up and 42 were face down. You are in a dark room holding the deck. How do you rearrange the deck into two subdecks so that they have the same number of cards facing up?

The cards can be flipped. So first, the specific numbers don’t matter. They don’t even have to be even numbers. The answer is simple.

Say there are \(n\) cards up in the full deck. If you just divide the deck without flipping cards, then however you divide, one subdeck will have \(m\in [0,n]\) cards up, and the other subdeck will have the complement \(n-m\) cards up. So take \(n\) cards out of the original deck as subdeck A, and flip them over. If this subdeck A had \(m\) cards up originally, now it has \(n-m\) cards up. In the remaining cards forming subdeck B, there were already \(n-m\) cards up. So the two subdecks have an equal number of cards up.

What is this “blog”

admin — Tue, 24 Oct 2006 16:23:51 +0000

…you speak of… what, do I write to myself? I only have 100MB.

First post and already TeX can be rendered. I stole the idea from fakalin.

\( \int_{0}^{1}\frac{x^{4}\left( 1-x\right) ^{4}}{1+x^{2}}dx = \frac{22}{7}-\pi \)

Some stuff » fakalin

a card problem

What is this “blog”