Binary Notation Card TrickDate: 12/10/2002 at 20:10:11 From: Jonathon Subject: Card Game My 7th grade teacher had a card game that had about 7 cards. She said she could guess your number by asking you if your number was on each card. Can you teach me how to play this game. All I remember is that one card had even numbers and the other had odd. Date: 12/11/2002 at 13:04:08 From: Doctor Ian Subject: Re: Card Game Hi Jonathan, I think I know the game you're talking about. I have a friend who pulls this all the time on unsuspecting children. To understand it, you have to know how binary ('base 2') notation works. Consider a number like 423. This is a shorthand representation for 4*100 + 2*10 + 3*1 That is, the value contributed by each digit is the digit times a power of 10: 10^2 10^1 10^0 ---- ---- ---- 4 2 3 = 4*100 + 2*10 + 3*1 3 0 9 = 3*100 + 0*10 + 9*1 and so on. This, by the way, is what makes zero so valuable. Instead of using all 10 digits in base 10, we can do the same sort of thing using only 2 digits (0 and 1), in base 2: 2^3 2^2 2^1 2^0 --- --- --- --- 0 0 0 1 = 0*8 + 0*4 + 0*2 + 1*1 = 1 0 0 1 0 = 1*2 = 2 0 0 1 1 = 1*2 + 1*1 = 3 0 1 0 0 = 1*4 = 4 0 1 0 1 = 1*4 + 1*1 = 5 0 1 1 0 = 1*4 + 1*2 = 6 0 1 1 1 = 1*4 + 1*2 + 1*1 = 7 1 0 0 0 = 1*8 = 8 and so on. Here are the binary representations of the numbers from 1 to 15: base 10 binary ------- ------- 1 0 0 0 1 2 0 0 1 0 3 0 0 1 1 4 0 1 0 0 5 0 1 0 1 6 0 1 1 0 7 0 1 1 1 8 1 0 0 0 9 1 0 0 1 10 1 0 1 0 11 1 0 1 1 12 1 1 0 0 13 1 1 0 1 14 1 1 1 0 15 1 1 1 1 Now we're ready to make our cards. To make the first card, we choose every number that has a '1' in the rightmost column: 1 3 5 7 9 11 13 15 To make our second card, we choose every number that has a '1' in the next column over (i.e., the second column from the right): 2 3 6 7 10 11 14 15 We make the next two cards the same way: 4 5 6 7 12 13 14 15 and 8 9 10 11 12 13 14 15 So here are our cards: Card 4 Card 3 Card 2 Card 1 +-----------+ +-----------+ +-----------+ +-----------+ | 8 9 10 11| | 4 5 6 7| | 2 3 6 7| | 1 3 5 7| |12 13 14 15| |12 13 14 15| |10 11 14 15| | 9 11 13 15| +-----------+ +-----------+ +-----------+ +-----------+ Now, let's pick a number at random, like 12. Which cards does it appear on? Cards 3 and 4. Let's add the first numbers that appear on those cards: 4 + 8 = 12. How about that! Let's try another number, like 7. It appears on cards 1, 2, and 3, so we'll add the first numbers from those cards: 1 + 2 + 4 = 7. Do you see what's going on? By telling you what cards a number appears on, I'm telling you the digits that appear in its binary representation. In other words, if the number 12 appears on cards 3 and 4, then its binary representation must be 1100. If the number 7 appears on cards 1, 2, and 3, its binary representation is 0111. And so on. And what is the first number on each card? It's the power of 2 that corresponds to the column we used to make the card! So, doing this: Choose a number from these cards. Find the cards that it appears on, and add the first numbers from those cards. The sum will be equal to the number you chose. +-----------+ +-----------+ +-----------+ +-----------+ | 8 9 10 11| | 4 5 6 7| | 2 3 6 7| | 1 3 5 7| |12 13 14 15| |12 13 14 15| |10 11 14 15| | 9 11 13 15| +-----------+ +-----------+ +-----------+ +-----------+ is _exactly_ the same as doing this: Choose a number from the following list. Find the columns next to it that contain a '1', and add the numbers at the tops of those columns. The sum will be equal to the number you chose. 8 4 2 1 ------- 1 0 0 0 1 2 0 0 1 0 3 0 0 1 1 4 0 1 0 0 5 0 1 0 1 6 0 1 1 0 7 0 1 1 1 8 1 0 0 0 9 1 0 0 1 10 1 0 1 0 11 1 0 1 1 12 1 1 0 0 13 1 1 0 1 14 1 1 1 0 15 1 1 1 1 When you add the first numbers of the cards on which a number appears, you're just adding up the binary representation of the number. You can do this with more numbers, of course. Adding one more card (one more column) will give you the numbers from 1 to 31; adding two columns will give you the numbers from 1 to 63; and so on. With N cards, you can use the numbers from 1 to (2^N-1). But it's just more numbers. The trick is exactly the same. Does this make sense? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/