2
$\begingroup$

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.

$\endgroup$
6
  • 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ This looks more like a math problem, not puzzle. $\endgroup$ – trolley813 Apr 9 '20 at 10:36
  • 4
    $\begingroup$ @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
7
$\begingroup$

Because

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

and

$2019=(4\times504)+3\equiv3\:(\text{mod}\;4)$,

we have

$3^{2019}=(3^4)^{504}\times3^3\equiv(1)^{504}\times3^3=27\equiv7\:(\text{mod}\;10)$,

so the answer is

$7$.

$\endgroup$
1
  • 4
    $\begingroup$ Equally interesting is: "what is the first digit of $3^{2019}$?". $\endgroup$ – WhatsUp Apr 13 '20 at 22:48
1
$\begingroup$

As we know,

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

As

$MOD(2019,4)=3$

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:

$7$

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.