It is fun to spend some time making arithmetic rules for different bases, though. Take hexadecimal for example. Sixteen digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A (10), B (11), C (12), D (13), E (14), and F (15). Therefore, the "tens" place is actually the "sixteens" place. For example,

*10*is 1 x 16 + 0, and

*21*is 2 x 16 +1. This means that you can count up to fifteen-16s in only two digits:

*FF*is 15 x 16 + 15 = 255. Thus, the "hundreds" place is actually the "256th-s" place. In hexadecimal,

*100*is 256. One of my favorite t-shirts, which my wife won't allow me to buy, reads, "1 in 100 people understand hexadecimal. The other two hundred fifty-five just stare at you." The place values continue the same pattern: the "ones" digit is multiplied by 16^0, or 1; the "tens" digit is multiplied by 16^1, or 16; the "hundreds" digit is multiplied by 16^2, or 256; the "thousands" digit is multiplied by 16^3, or 4096; and so on. Now we can start to make an addition table!

Does our standard old algorithm for adding multidigit numbers work? Let's try adding

*1C*+

*59*. First, tackle

*C*+

*9*. Our table says the answer is

*15*. Put the

*5*down, carry the

*1*. Now in the sixteens place we have the carried

*1*plus

*1*+

*5*, so

*7*. Our final answer is

*75*. Is this right? To check, convert everything into decimal.

*1C*is 28,

*59*is 89, and

*75*is 117. And yes, that's correct, so the standard old algorithm works regardless of the base of the number system.

What about multiplication? First, we need a multiplication table:

Now we can try our standard old algorithm for multiplication. Let's do

*1C*x

*59*. First, separate it into

*1C*x

*9*and

*1C*x

*50*. So,

*C*x

*9*is

*6C*, from the table. I write down

*C*in the ones place and carry the

*6*.

*1*x

*9*is

*9*, then add the

*6*, so

*9*+

*6*is

*F*. This means

*1C*x

*9*=

*FC*. I repeat the process for

*1C*x

*50*, making sure to write down a 0 in the ones place. This yields

*1C*x

*50*=

*8C0*. Now, the final step is to add

*FC*and

*8C0*. I get

*9BC*, so my final answer is

*1C*x

*59*=

*9BC*. Is this correct? To check, convert everything into decimal.

*1C*is 28,

*59*is 89, and

*9BC*is 2492. And yes, that's correct, so the standard old algorithm for multiplication works regardless of the base of the number system. All of this is pretty darn neat.

Here is a great trick I got from Cathy Varner for binary. Binary works with only two digits, 0 and 1, so 2 is represented as "10"; i.e., 1 * 2^1 + 0 * 2^0. Every number can be represented by adding some combination of the powers of two: for example, 11 is represented as "1011" -- 1 * 2^3 + 0 * 2^2 + 1 * 2^1 + 1 * 2^0. The trick is to ask students to think of their birthday (the day of the month, 1-31), then use a series of cards to figure out the binary representation of their birthday, and then "guess" it correctly every time. For example, the first card includes all the odd numbers (1, 3, 5, etc.). If the student says "yes, my birthday is on this card," I know the ones digit, the 2^0 digit, is "1", not "0". So in this case, I keep a +1 in my head. The second card includes every number where the second digit in binary, the 2^1 digit, is "1", not "0". So if the student gives a "yes, my birthday is on that card," I keep a +2 in my head. I also have cards for the 2^2 digit (+4), the 2^3 digit (+8), and the 2^4 digit (+16). Since 2^5 = 32, I do not need this card. The highest birthday number is 31, which in binary is "11111" -- that is, +1 +2 +4 +8 +16.

All you have to do is make the correct cards. This table should help:

This shows you what numbers to put on each card. The students answer yes or no, and all you need to do is keep a running tally in your head. Making the card is the only challenging part!

I came up with a powers of three trick. I ask them to think of a number from 0-26: age of a sibling, perhaps. I show them these three cards, and ask whether the number is present in black, in red, or not present. Can you see how it works?

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