Short Trick Number System

Apr 2018
6
0
India
Hello friends ,
you can learn a lot in 3 minutes
[youtube]mMQO7f_GHa0[/youtube]
 

topsquark

Math Team
May 2013
2,530
1,050
The Astral plane
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
 

Denis

Math Team
Oct 2011
14,592
1,026
Ottawa Ontario, Canada
This post (which is your 2nd post) is as useless as your 1st post...
 
May 2016
1,310
551
USA
Subscribe. Spam. Ban the jerk.
 
Apr 2018
6
0
India
This video is a trick to find unit digit of very big multiplications.
 
Aug 2012
2,496
781
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?
 

Denis

Math Team
Oct 2011
14,592
1,026
Ottawa Ontario, Canada
...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.
 
Apr 2018
6
0
India
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
 
Apr 2018
6
0
India
...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
 
Oct 2009
942
367
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 $2^5 = 2$.
 
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