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 December 9th, 2017, 05:01 AM #1 Newbie   Joined: Mar 2017 From: Connecticut Posts: 8 Thanks: 0 Multiplication Algorithm Hello all, I made a post a few months back asking about methods for squaring 2-digit numbers after I recognized a pattern and you friendly folks told me I had actually stumbled upon an old method that had long fallen out of use. My question today focuses on multiplication in general. Last night, I was mindlessly doing 9 times tables when I realized that 9 times a number is the same as 10 times a number minus that number. For instance, I saw that 9x5 = 10x5 - 5. So then I tried it with a few other numbers like the 8 times table and the 7 times table (with an added step, as shown below) and found the same to be true. All in all, I wrote out the algorithm like this: (X)(Y) = 10Y - (10-X)Y E.g.: (7)(15) = (10)15 - (10-7)15 =150 - (3)15 =150 - 45 = 105 I know that this method is probably not really useful and I still haven't tested it with a wide enough range of numbers to see if it's always true. However, I'm still curious as to whether anyone recognizes this as an actual method or not. Thanks!
December 9th, 2017, 05:47 AM   #2
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Quote:
 Originally Posted by burner830 Hello all, I made a post a few months back asking about methods for squaring 2-digit numbers after I recognized a pattern and you friendly folks told me I had actually stumbled upon an old method that had long fallen out of use. My question today focuses on multiplication in general. Last night, I was mindlessly doing 9 times tables when I realized that 9 times a number is the same as 10 times a number minus that number. For instance, I saw that 9x5 = 10x5 - 5. So then I tried it with a few other numbers like the 8 times table and the 7 times table (with an added step, as shown below) and found the same to be true. All in all, I wrote out the algorithm like this: (X)(Y) = 10Y - (10-X)Y E.g.: (7)(15) = (10)15 - (10-7)15 =150 - (3)15 =150 - 45 = 105 I know that this method is probably not really useful and I still haven't tested it with a wide enough range of numbers to see if it's always true. However, I'm still curious as to whether anyone recognizes this as an actual method or not. Thanks!
Burner, you have not said how old you are or what you are studying. I am guessing that you are studying arithmetic and are about 10. My answer is going to assume that to be the case so I apologize in advance if my answer is a bit simplistic.

It is a basic rule of arithmetic that if a, b, and c are any numbers, then

$a \times (b + c) \equiv (a \times b) + (a \times c) \text { and } a \times (b - c) \equiv (a \times b) - ( \times c).$

This is called the distributive law. The funny looking equal sign means that the equality is always true.

So let's look at your algorithm generally.

$z = 10y - (10 - x)y \iff z = 10y - (10y - xy) \iff z = (10y - 10y) + xy \iff z = xy.$

It will always be true. Long ago, before hand calculators, people who had to do arithmetic a lot used your trick and others like them to do mental arithmetic.

$28 \times 30 = 30 \times (30 - 2) = 900 - 60 = 840.$

$32 \times 28 = (30 + 2) \times (30 - 2) = 30^2 - 2^2 = 900 - 4 = 896.$

You said in your previous thread that you are not good at math. I very much doubt that to be true. I suspect that you find arithmetic boring and so your attention wanders and you make silly mistakes. Arithmetic IS boring, but the further you go in math, the more and more interesting it becomes. Stop persuading yourself that you are not good at math. Just recognize that computation punishes carelessness.

December 9th, 2017, 01:07 PM   #3
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 Originally Posted by JeffM1 Burner, you have not said how old you are or what you are studying. I am guessing that you are studying arithmetic and are about 10. My answer is going to assume that to be the case so I apologize in advance if my answer is a bit simplistic. It is a basic rule of arithmetic that if a, b, and c are any numbers, then $a \times (b + c) \equiv (a \times b) + (a \times c) \text { and } a \times (b - c) \equiv (a \times b) - ( \times c).$ This is called the distributive law. The funny looking equal sign means that the equality is always true. So let's look at your algorithm generally. $z = 10y - (10 - x)y \iff z = 10y - (10y - xy) \iff z = (10y - 10y) + xy \iff z = xy.$ It will always be true. Long ago, before hand calculators, people who had to do arithmetic a lot used your trick and others like them to do mental arithmetic. $28 \times 30 = 30 \times (30 - 2) = 900 - 60 = 840.$ $32 \times 28 = (30 + 2) \times (30 - 2) = 30^2 - 2^2 = 900 - 4 = 896.$ You said in your previous thread that you are not good at math. I very much doubt that to be true. I suspect that you find arithmetic boring and so your attention wanders and you make silly mistakes. Arithmetic IS boring, but the further you go in math, the more and more interesting it becomes. Stop persuading yourself that you are not good at math. Just recognize that computation punishes carelessness.
Thanks for the reply but, um, I'm definitely not 10... Which, if my observation leads you to believe I'm functioning at a 5th grade level, would explain the whole "I'm not a math genius" thing. Also not sure what you meant by making silly mistakes but whatevs.

All I wanted to know was if this was an actual way to do multiplication in the past which you did answer for me. So thanks.

December 9th, 2017, 01:16 PM   #4
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 Originally Posted by burner830 All I wanted to know was if this was an actual way to do multiplication in the past which you did answer for me. So thanks.
I had a roommate that used to impress folks with his ability to multiply quickly in his head. He used this exact trick.

So yeah, it's been known and used in the past.

December 9th, 2017, 02:13 PM   #5
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Quote:
 Originally Posted by JeffM1 I am guessing that you are studying arithmetic and are about 10.
Yes but in what base?

December 9th, 2017, 09:18 PM   #6
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 Originally Posted by burner830 Thanks for the reply but, um, I'm definitely not 10... Which, if my observation leads you to believe I'm functioning at a 5th grade level, would explain the whole "I'm not a math genius" thing. Also not sure what you meant by making silly mistakes but whatevs. All I wanted to know was if this was an actual way to do multiplication in the past which you did answer for me. So thanks.
You give no information about yourself so I must guess. I apologized in advance if my guess was incorrect. When someone says that he or she is contemplating the mysteries of the nine's table, I do not conclude that he or she is deep into abstract algebra. Given your sensitivities, I shall remember in the future to ignore your posts.

December 10th, 2017, 04:02 AM   #7
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 Originally Posted by JeffM1 You give no information about yourself so I must guess. I apologized in advance if my guess was incorrect. When someone says that he or she is contemplating the mysteries of the nine's table, I do not conclude that he or she is deep into abstract algebra. Given your sensitivities, I shall remember in the future to ignore your posts.
I wouldn't say mindlessly doing 9 times tables equates to contemplating the mysteries of it. And I wouldn't say that correcting you about a big assumption on my age is being "sensitive" and calls for the ignoring of my future questions and discussions which are wholly unrelated to age in the first place... But ok.

December 10th, 2017, 04:04 AM   #8
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 Originally Posted by romsek I had a roommate that used to impress folks with his ability to multiply quickly in his head. He used this exact trick. So yeah, it's been known and used in the past.
Ah, thank you very much! Good to know!

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