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 May 12th, 2018, 07:08 AM #1 Newbie   Joined: Apr 2018 From: India Posts: 6 Thanks: 0 Short Trick Number System Hello friends , you can learn a lot in 3 minutes
 May 12th, 2018, 07:27 AM #2 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,878 Thanks: 760 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
 May 12th, 2018, 08:10 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,102 Thanks: 907 This post (which is your 2nd post) is as useless as your 1st post...
 May 12th, 2018, 03:12 PM #4 Senior Member   Joined: May 2016 From: USA Posts: 1,117 Thanks: 462 Subscribe. Spam. Ban the jerk.
 May 12th, 2018, 07:19 PM #5 Newbie   Joined: Apr 2018 From: India Posts: 6 Thanks: 0 This video is a trick to find unit digit of very big multiplications.
May 13th, 2018, 09:22 AM   #6
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Quote:
 Originally Posted by davidJ 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?

 May 13th, 2018, 09:55 AM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,102 Thanks: 907 ...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.
May 13th, 2018, 07:18 PM   #8
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Quote:
 Originally Posted by Maschke 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

May 13th, 2018, 07:18 PM   #9
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Quote:
 Originally Posted by Denis ...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

May 13th, 2018, 11:54 PM   #10
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Quote:
 Originally Posted by davidJ 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 $2^5 = 2$.

Last edited by Micrm@ss; May 14th, 2018 at 12:05 AM.

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