Which would be the last digit for $3^{2019}$ ?

You can

check last digits for $3^x$ with $x=\{0,1,2,3,4,5,6\}$ and see if something is repeated.

And afterwards

think about modulus for the exponent number and the pattern found.

  • 8
    $\begingroup$ No need to include hints, this is easy enough already for someone who knows maths :-) $\endgroup$ – Rand al'Thor Apr 9 '20 at 9:34
  • $\begingroup$ Should I remove the hints? Just new to this Stackexchange site and not sure when and whether should I include hints when knowing the answer. $\endgroup$ – Cedric Zoppolo Apr 9 '20 at 10:06
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    $\begingroup$ Hints are usually added after posting the puzzle, as a way to point people in the right direction if nobody gets the answer for a while. I'd say you can remove them. And yes, LaTeX format is also possible in titles, but it makes questions ineligible for the Hot Network Questions list. $\endgroup$ – Rand al'Thor Apr 9 '20 at 10:11
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    $\begingroup$ This looks more like a math problem, not puzzle. $\endgroup$ – trolley813 Apr 9 '20 at 10:36
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    $\begingroup$ @trolley813 I disagree. It's easy for those of us who've studied some number theory, sure, but the method of solution would be very interesting and "aha"-ish for someone who hasn't seen it before. I could easily imagine this as an olympiad problem, for example (not IMO but maybe a subnational olympiad). Sometimes we forget that what's second nature to us may be a fascinating innovation for non-mathematicians :-) $\endgroup$ – Rand al'Thor Apr 9 '20 at 11:10


$3^4=81\equiv1 \:(\text{mod}\;10)$,



we have


so the answer is


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    $\begingroup$ Equally interesting is: "what is the first digit of $3^{2019}$?". $\endgroup$ – WhatsUp Apr 13 '20 at 22:48

As we know,

Powers of 3 are numbers ending in $\{1,3,9,7\}$ sequentially.



So we know that

The last digit for $3^{2019}$ will be the forth in the sequence stated in the beginning. As if result was 0 it would have been the first element.

So the result is that the last digit for $3^{2019}$ is:


  • $\begingroup$ Well, the first spoilerblock ("As we know ...") is really the key point. Rather than assuming such a pattern (even though I do know it's always going to be like that), in my answer I gave an argument which anyone could follow even knowing nothing about modular arithmetic, as long as they know what $\equiv\:(\text{mod}\;n)$ actually means. $\endgroup$ – Rand al'Thor Apr 9 '20 at 10:29
  • $\begingroup$ The list in the first spoilerblock isn’t in the correct order, although it’s not specified if it’s 0-indexed or 1-indexed $\endgroup$ – Charlie Harding Apr 16 '20 at 19:45
  • $\begingroup$ @CharlieHarding, you are right. Just corrected the answer. Thanks. $\endgroup$ – Cedric Zoppolo Apr 16 '20 at 23:13

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