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May 12th, 2018, 08:08 AM   #1
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Short Trick Number System

Hello friends ,
you can learn a lot in 3 minutes
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May 12th, 2018, 08:27 AM   #2
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Math Focus: Wibbly wobbly timey-wimey stuff.
Part of my problem is that I don't speak the language and can't follow it. An example of my problem is this: At 0:36 the video seems to be claiming that says $\displaystyle 656 \times 838 \times 972$ is somehow 6? My guess is that we are looking to find the last digit in the multiplication?

-Dan
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May 12th, 2018, 09:10 AM   #3
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This post (which is your 2nd post) is as useless as your 1st post...
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May 12th, 2018, 04:12 PM   #4
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Subscribe. Spam. Ban the jerk.
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May 12th, 2018, 08:19 PM   #5
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This video is a trick to find unit digit of very big multiplications.
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May 13th, 2018, 10:22 AM   #6
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Quote:
Originally Posted by davidJ View Post
This video is a trick to find unit digit of very big multiplications.
Didn't watch the vid but isn't the unit digit just the product of the unit digits of the factors, reduced mod 10? For example the unit digit of 53453543543543 x 543543535435436 is 8. Right?
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May 13th, 2018, 10:55 AM   #7
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...and why d'hell is "very big" scary?
You really end up multiplying a few digits.
Plus no multiplication required if one or more of those last digits is 0 or 5.
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May 13th, 2018, 08:18 PM   #8
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Quote:
Originally Posted by Maschke View Post
Didn't watch the vid but isn't the unit digit just the product of the unit digits of the factors, reduced mod 10? For example the unit digit of 53453543543543 x 543543535435436 is 8. Right?
It became a problem when you have powers on your dgits.
Watch it complete you will understand
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May 13th, 2018, 08:18 PM   #9
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Quote:
Originally Posted by Denis View Post
...and why d'hell is "very big" scary?
You really end up multiplying a few digits.
Plus no multiplication required if one or more of those last digits is 0 or 5.
This video also contains digits with powers, what do you do in those cases
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May 14th, 2018, 12:54 AM   #10
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Quote:
Originally Posted by davidJ View Post
This video also contains digits with powers, what do you do in those cases
Euler's theorem. Which you apparently either didn't apply or didn't mention.
It states that if $a$ is not divisible by $2$ or $5$, then the last digit of $a^4 = 1$.
Thus for example, to compute
$$333^{4323133}$$
we write $4323133 = 4*1080783 + 1$
Hence in mod 10
$$333^{4323133} = (333^4)^{1080783}333 = 333 = 3.$$
Isn't this a lot easier?????

Now if $a$ is divisible by $5$, then the last digit of $a$ is always $5$ or $0$, and it is easy to see which.
If $a$ is divisible by $2$, then use that the last digit of $2^5 = 2$.
Thanks from Denis, topsquark and davidJ

Last edited by Micrm@ss; May 14th, 2018 at 01:05 AM.
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