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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.

Cedric Zoppolo
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    No need to include hints, this is easy enough already for someone who knows maths :-) – Rand al'Thor Apr 09 '20 at 09:34
  • 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. – Cedric Zoppolo Apr 09 '20 at 10:06
  • @Randal'Thor Also, I see you edited my question and added LateX format. Is that possible on titles as well? – Cedric Zoppolo Apr 09 '20 at 10:07
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    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. – Rand al'Thor Apr 09 '20 at 10:11
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    This looks more like a math problem, not puzzle. – trolley813 Apr 09 '20 at 10:36
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    @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 :-) – Rand al'Thor Apr 09 '20 at 11:10

3 Answers3

9

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

$7$.

Rand al'Thor
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1

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:

$7$

Cedric Zoppolo
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  • 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. – Rand al'Thor Apr 09 '20 at 10:29
  • 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 – Charlie Harding Apr 16 '20 at 19:45
  • @CharlieHarding, you are right. Just corrected the answer. Thanks. – Cedric Zoppolo Apr 16 '20 at 23:13
0

The list of last digits:

$3$

$9$ ($3×3$)

$7$ ($3×3×3$)

$1$ ($3×3×3×3$)

And then the cycle repeats again. So, 3 to the 2019 th power is same as $3^3$(There is a cycle of 4). Which gives 27,which have a last digit of 7.

A Math guy
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