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Re: 8th Grade Examination 1912. Micro "Can you tell me what 57 MULTIPLIED by 135 is?"

By: ribit in 6TH POPE | Recommend this post (0)
Sat, 03 Sep 22 8:58 PM | 16 view(s)
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Msg. 35206 of 58662
(This msg. is a reply to 35201 by Fiz)

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kids with at least average intelligence could be taught to EASILY *compute* all their times tables above 2x.

...I got a masters degree in times tables in Coach Greer's detention class every afternoon after all the Slackers with no desire to enhance their abilities had gone home.




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Liberals are like a "Slinky". Totally useless, but somehow ya can't help but smile when you see one tumble down a flight of stairs!




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Re: 8th Grade Examination 1912. Micro "Can you tell me what 57 MULTIPLIED by 135 is?"
By: Fiz
in 6TH POPE
Sat, 03 Sep 22 7:21 PM
Msg. 35201 of 58662

Hi Micro,

Yes, I used to use a slide rule. Now I try to use my head:

You might enjoy watching the short movie excerpt from "Gifted", first:

http://www.youtube.com/watch?v=37meAwQqPsE

"Can you tell me what 57 MULTIPLIED by 135 is?"

So here is the way I do 57 x 135 without paper. It hinges on the fact that any two numbers can be restated as (C+a) and (C+b), respectively, and (C+a)(C+b) = C(C+a+b)+ a*b. (FOIL and factor) Also, 135 x57=3[45x57] (1)
so, choosing C=50, 45x57=50(50+7-5)+(7)(-5) =(100/2)(52) +(-35). (2)
=2600-35=2565.
∴3[2565]=7500+195=7695.
Check: Casting out nines we get 7695->0 and 135->0 and 57->3. 0=(0)3. YES. It is correct.

6x7? 10(3)+12=42. Or, if we choose 5 as a Center, 5(Cool+1*2=42.

With slight variations this is a general strategy which can be used to "figure out" pretty nearly any multiplication where the multiplicands are relatively close together (e.g., 113 x 107; w C=100; 17x13 w C=10). Note 50=100/2 and 20=10*2, etc. and the range expands easily! When multiplicands are further from C, other algebra tricks can be used. You get the answer, and you exercise your algebra to boot.

Let's try it again with a much simpler problem, but still one you might think it would take a genius to solve: 17x13.
choose C=10. ∴17x13 = 20(10) +21=221. Check? 5?=8*4->32->5? YES. It is correct.

The truth is, kids with at least average intelligence could be taught to EASILY *compute* all their times tables above 2x. Once they learn this mathematical strategy, the whole realm of numbers and math would be dramatically opened to them. They would start to get curious, as their confidence rightly goes up, and they would discover other patterns, which could be expressed in terms of strategy and algorithm.

This also naturally segues in to algebra, bringing math alive. (x+a)(y+b) or (x-a)(y-b) would have common applications and more intuitive meaning. 37^2? 1200+169=1369 comes out naturally from using 25 as Center(4) and noting that 50=2x25. Check: 1vs1? YES.

Note, I deliberately didn't explain all the sub-strategies and rationale, in the (C+a)(C+b) = C(C+a+b)+ a*b derivation and application. I also didn't get into the sub-strategy and of "casting out nines" as a fast accuracy check. Not only would a complete explication distract from the point, it would also occlude the fact that I use sub-strategies and most of them are now automatic (almost unconscious). It would also make the underlying simplicity, much harder to follow here.

FWIW, I only started learning these "basic" algebra applications a few years ago. Now the recognition and application of patterns keeps expanding. Eventually, I expect I could just turn everything over to my unconscious and just come back for the answer. I'm not there yet. What if I was taught as a 1st grader? What about teaching every first grader? Start with small numbers and grow up! Inch by inch, math is a cinch!

P.S. I thought you might enjoy this given your background and the topic. I try to do a problem or three a day, just to exercise my brain -- and give me solid assurance that my brain is still pretty functional! I assure you it is easier than it seems and actually rather enjoyable when you do something rather difficult or get a new insight into how or why something can be made to work!

I use some other algebraically provable tricks to find squares of numbers, using the relationship between 25, 50, and 100. So 42^2= 100(42-25)+(50-42)^2. =1700+8^2==1764.

Check? removing 9s from 1764, I just start adding up the digits and dropping 9 whenever I come to it. So...17=>8; 64=>10 and 8+10=>1 8=>99=>0. That means the multiplicands need to reduce to 0. Now, 42=>6. So 6*6=>36=>9=>0. So, better than 90% chance you have the right answer. Check! Casting out 9s is just a function which depends on the fact that, ultimately, 42 x 42 is just another way of writing 1764!


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