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May 14th, 2018, 06:39 AM   #11
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 Originally Posted by Micrm@ss 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 $2^5 = 2$.

This is the same thing explained in this video but in a better way May 14th, 2018, 08:07 AM #12 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Define "better way". May 14th, 2018, 09:27 AM   #13
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Quote:
 Originally Posted by davidJ This is the same thing explained in this video but in a better way
Sure except that you make it appear like a magic trick, while I explain why it works. Tags number, short, system, trick Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post johnbannister2713 New Users 1 July 4th, 2017 11:06 PM HawkI Number Theory 13 November 26th, 2016 03:18 AM myrv Algebra 6 January 30th, 2010 05:11 PM dbboy Number Theory 1 April 12th, 2008 08:14 PM

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