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# Or like a computer. Or like an Egyptian computer.

I'm confused. Isn't 35*45 [x]5 * [(x+2)]5 where x=7?

And I honestly don't see how this could possibly be true.

[x]5 * [(x+1)]5 = 75*x(x+2)

25*(x)(x+1) = 75*x(x+2)

25(x^2+x)=75(x^2+2x)

x^2+x=3x^2+2x

For x = 7

49+7 = 3*49 + 14

Surely I'm missing something?

Using Crash's tip for squares of 5s, you can come up with formulas for numbers ending in other digits as well.

For example, for two digit numbers ending in 1, the square is 100x*x + 20x + 1, so for example:

posted by Deathalicious at 7:09 AM on December 10, 2010

Post

# Or like a computer. Or like an Egyptian computer.

December 9, 2010 6:37 AM Subscribe

I love math. I wish my early math teachers had also loved math.

posted by odinsdream at 6:47 AM on December 9, 2010 [1 favorite]

posted by odinsdream at 6:47 AM on December 9, 2010 [1 favorite]

Don't be silly. The Egyptians didn't know math(s). Those buildings were built by aliens.

posted by clvrmnky at 6:51 AM on December 9, 2010

posted by clvrmnky at 6:51 AM on December 9, 2010

Multiply like an alien.

posted by overeducated_alligator at 6:52 AM on December 9, 2010 [1 favorite]

posted by overeducated_alligator at 6:52 AM on December 9, 2010 [1 favorite]

All the old paintings on the tomb they do binary, donchaknow ...

posted by bwg at 7:04 AM on December 9, 2010

posted by bwg at 7:04 AM on December 9, 2010

Cool, thank you!

And - the on/off icon comes from the state of electrical flow in early computer switches? - Neat!

posted by carter at 7:14 AM on December 9, 2010

And - the on/off icon comes from the state of electrical flow in early computer switches? - Neat!

posted by carter at 7:14 AM on December 9, 2010

Ancient Egyptian Science: A Source Book, Vol. Three Ancient Egyptian Mathematics

Egyptian numerals

Mathematics in Egyptian Papyri

History Topics: Ancient Egyptian mathematics

Also interesting is Assyro-Babylonian mathematics which uses a base 60 numeral system.

Babylonian Mathematics

posted by AElfwine Evenstar at 7:15 AM on December 9, 2010 [2 favorites]

Egyptian numerals

Mathematics in Egyptian Papyri

History Topics: Ancient Egyptian mathematics

Also interesting is Assyro-Babylonian mathematics which uses a base 60 numeral system.

Babylonian Mathematics

posted by AElfwine Evenstar at 7:15 AM on December 9, 2010 [2 favorites]

Actually, I'm pretty sure most computers do this. Still a neat technique though.

posted by Chuckles at 7:33 AM on December 9, 2010

posted by Chuckles at 7:33 AM on December 9, 2010

So, now I understand how binary works. thanks!

posted by oddman at 7:40 AM on December 9, 2010 [3 favorites]

posted by oddman at 7:40 AM on December 9, 2010 [3 favorites]

2094 / 13 GO!

posted by From Bklyn at 8:05 AM on December 9, 2010

posted by From Bklyn at 8:05 AM on December 9, 2010

"No more tears for 3rd graders" yeah! Instead of memorizing 6x7=42 all they have to do is be able to double 347 in their heads (twice) while also remembering that 6 is 2+4! It's so simple!

Computers use binary because it's natural to the switch paradigm. The Egyptians used binary (I'm guessing) because doing math base 60 isn't too convenient in terms of the number of symbols you have to remember, not to mention an enormous multiplication table (3600 1 digit products!).

But if your numbers are already in base 10, it's much more natural to use this same algorithm in that same base.

17 x 25 is 10 x 25 + 7 x 25.

10 x 25 is easy: 250

7 x 25 is pretty easy, but if not keep in mind that 7 = 4 + 3.

4 x 25 = 100

3 x 25 = 75

So we have 250 + 100 + 75 = 425.

The is the exact same algorithm but you don't have to convert bases (double) before hand NOR do you have to remember the binary representation of the other number because the base 10 digits are sitting there right in front of you.

posted by DU at 8:15 AM on December 9, 2010 [3 favorites]

Computers use binary because it's natural to the switch paradigm. The Egyptians used binary (I'm guessing) because doing math base 60 isn't too convenient in terms of the number of symbols you have to remember, not to mention an enormous multiplication table (3600 1 digit products!).

But if your numbers are already in base 10, it's much more natural to use this same algorithm in that same base.

17 x 25 is 10 x 25 + 7 x 25.

10 x 25 is easy: 250

7 x 25 is pretty easy, but if not keep in mind that 7 = 4 + 3.

4 x 25 = 100

3 x 25 = 75

So we have 250 + 100 + 75 = 425.

The is the exact same algorithm but you don't have to convert bases (double) before hand NOR do you have to remember the binary representation of the other number because the base 10 digits are sitting there right in front of you.

posted by DU at 8:15 AM on December 9, 2010 [3 favorites]

I learned about this in a great book about non-western science and math called "Lost Discoveries"- it's a great read if anyone is interested.

posted by rmless at 8:21 AM on December 9, 2010 [1 favorite]

posted by rmless at 8:21 AM on December 9, 2010 [1 favorite]

Yeah, this should look familiar - it's exactly the long multiplication you learned in gradeschool, but performed in base-2 rather than base-10. Notice how clean everything looks in an alphabet of only two symbols.

If you're new to math and got a kick out of this, try proving to yourself that any natural number (counting numbers like 6,7,8, 365, 1001, and so on, as large as you can imagine) can be expressed by adding together nothing but powers of 2.

Take 365.

= 256 + 64 + 32 + 16 + 4 + 2 + 1.

= 2^8 + 2^6 + 2^5 + 2^4 + 2^2 + 2^1 + 2^0. (Remember that 2^0 = 1.) All powers of 2.

Try to conceive of a natural number that

posted by kid ichorous at 8:46 AM on December 9, 2010

If you're new to math and got a kick out of this, try proving to yourself that any natural number (counting numbers like 6,7,8, 365, 1001, and so on, as large as you can imagine) can be expressed by adding together nothing but powers of 2.

Take 365.

= 256 + 64 + 32 + 16 + 4 + 2 + 1.

= 2^8 + 2^6 + 2^5 + 2^4 + 2^2 + 2^1 + 2^0. (Remember that 2^0 = 1.) All powers of 2.

Try to conceive of a natural number that

*isn't*expressible as a sum of powers of two. What would have to be true about this number?posted by kid ichorous at 8:46 AM on December 9, 2010

I vote for DU's method. Doing base 2 in my head isn't as intuitive. For example, to write out 365 as kid ichorous does, I first write out the powers of 2 (performing simple multiplications in my head), then start doing basic algebra (365 is greater than 256, so 256 gets a 1. Now I subtract 256 from 365, take the remainder, and go on to the next digit, etc...). All that work adds up to be more complex than the initial base 10 calculations.

What I like is simple math formulas that allow you to "cheat". Here's two good examples of what I'm talking about:

Any number ending in 5 squared can be abbreviated to x * x+1, append 25. So 35 squared is 3*4 = 12, append 25 = 1225. 55 squared is 5*6, append 25 = 3025, etc...

Another example: 9 multiplication. 9*x = x-1, append 9-(x-1). Examples:

9 * 8 = (8-1) append (9-7) = 72

( * 3 = (3 - 1) append (9 - 2) = 27

These are pretty easy to remember, and easy to prove if you want to verify they always work. Anyone have more?

posted by Crash at 10:01 AM on December 9, 2010 [1 favorite]

What I like is simple math formulas that allow you to "cheat". Here's two good examples of what I'm talking about:

Any number ending in 5 squared can be abbreviated to x * x+1, append 25. So 35 squared is 3*4 = 12, append 25 = 1225. 55 squared is 5*6, append 25 = 3025, etc...

Another example: 9 multiplication. 9*x = x-1, append 9-(x-1). Examples:

9 * 8 = (8-1) append (9-7) = 72

( * 3 = (3 - 1) append (9 - 2) = 27

These are pretty easy to remember, and easy to prove if you want to verify they always work. Anyone have more?

posted by Crash at 10:01 AM on December 9, 2010 [1 favorite]

For 5s, [x]5 * [(x+1)]5 = [x(x+2)]75. i.e. 35*45 = 1575, 55*65 = 3575, etc.

9 is a useful crude checksum for addition and multiplication. First, realize that the remainder of a number when divided by 9 (modulo 9, in other words) is the same as the remainder of the sum of the digits of the number when divided by 9. In other words, 3751 mod 9 is the same as (3+7+5+1) mod 9 = 16 mod 9 = 7 mod 9. As a checksum, that operation (adding up the digits mod 9, let's call it α) is preserved by addition and multiplication. In other words, for any two intergers a and b

α(a+b) = α(a) + α(b) (mod 9)

and

α(a*b) = α(a) * α(b) (mod 9)

So, let's say you need to add 572+723. You come up with 1295. Are you correct? A basic checksum here would be that α(572) = (5+7+2) mod 9 = 5 mod 9 and α(723) = (7+2+3) mod 9 = 3 mod 9, so we hope that &alpha(1295) is 3+5 mod 9 = 8 mod 9. And in fact (1+2+9+5) mod 9 is equal to 8. Similary, for multiplication, we would know that the product's α value would have to be 5*3 mod 9 = 6 mod 9. And in fact α(413556) = (4+1+3+5+5+6) mod 9 = 24 mod 9 = 6 mod 9.

Now, obviously this is a very crude checksum. If your calculations were off by some multiple of 9, it wouldn't catch it. But it should help eliminate a good chunk of off-by-one carrying errors and the like.

posted by kmz at 10:59 AM on December 9, 2010

9 is a useful crude checksum for addition and multiplication. First, realize that the remainder of a number when divided by 9 (modulo 9, in other words) is the same as the remainder of the sum of the digits of the number when divided by 9. In other words, 3751 mod 9 is the same as (3+7+5+1) mod 9 = 16 mod 9 = 7 mod 9. As a checksum, that operation (adding up the digits mod 9, let's call it α) is preserved by addition and multiplication. In other words, for any two intergers a and b

α(a+b) = α(a) + α(b) (mod 9)

and

α(a*b) = α(a) * α(b) (mod 9)

So, let's say you need to add 572+723. You come up with 1295. Are you correct? A basic checksum here would be that α(572) = (5+7+2) mod 9 = 5 mod 9 and α(723) = (7+2+3) mod 9 = 3 mod 9, so we hope that &alpha(1295) is 3+5 mod 9 = 8 mod 9. And in fact (1+2+9+5) mod 9 is equal to 8. Similary, for multiplication, we would know that the product's α value would have to be 5*3 mod 9 = 6 mod 9. And in fact α(413556) = (4+1+3+5+5+6) mod 9 = 24 mod 9 = 6 mod 9.

Now, obviously this is a very crude checksum. If your calculations were off by some multiple of 9, it wouldn't catch it. But it should help eliminate a good chunk of off-by-one carrying errors and the like.

posted by kmz at 10:59 AM on December 9, 2010

*For 5s, [x]5 * [(x+1)]5 = [x(x+2)]75. i.e. 35*45 = 1575, 55*65 = 3575, etc.*

I'm confused. Isn't 35*45 [x]5 * [(x+2)]5 where x=7?

And I honestly don't see how this could possibly be true.

[x]5 * [(x+1)]5 = 75*x(x+2)

25*(x)(x+1) = 75*x(x+2)

25(x^2+x)=75(x^2+2x)

x^2+x=3x^2+2x

For x = 7

49+7 = 3*49 + 14

Surely I'm missing something?

Using Crash's tip for squares of 5s, you can come up with formulas for numbers ending in other digits as well.

For example, for two digit numbers ending in 1, the square is 100x*x + 20x + 1, so for example:

31^2 900 60 1 --- 961 71 ^ 2 4900 140 1 ---- 5041Not nearly as elegant as the squares for 5.

posted by Deathalicious at 7:09 AM on December 10, 2010

Ohhhhhhh never mind. x is a digit, not a variable being multiplied by 5.

D'oh.

posted by Deathalicious at 7:11 AM on December 10, 2010

D'oh.

posted by Deathalicious at 7:11 AM on December 10, 2010

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