What is the last digit of $3^{2011}$

Question

The question speaks for itself, since the question comes from a contest environment, one where the use of calculators is obviously not allowed, can anyone perhaps supply me with an easy way to calculate the first of last digit of such situations?

My intuition said that I can look at the cases of $3$ with an index ending in a $1$, so I looked at $3^1=3$, $3^{11}=177,147$ and $3^{21}= 10,460,353,203$. So there is a slight pattern, but I'm not sure if it holds, and even if it does I will have shown it holds just for indices with base $3$, so I was wondering whether there is an easier way of knowing.

Any help is appreciated, thank you.

@Dr.SonnhardGraubner It is not very helpful to repeat the question. — amsmath, Nov 05 '17 at 17:52
See https://math.stackexchange.com/questions/679842/find-the-last-two-digits-of-345 and https://math.stackexchange.com/questions/789050/last-3-digits-of-3999 — lab bhattacharjee, Nov 05 '17 at 18:05

score 4 · Answer 1 · answered Nov 05 '17 at 17:37

4

Hint: Consider the first few powers of $3$ (say, the first five) and look for a pattern.

answered Nov 05 '17 at 17:37

Aaron Montgomery

10,688

score 3 · Accepted Answer · 2017-11-05T17:50:26.057

3

Actually 3^1=3 , 3^2=9 ,3^3=27 ,3^4=81 . These are the only four numbers that come at units place to powers of 3. So to find any last digit of 3^2011 divide 2011 by 4 which comes to have 3 as remainder . Hence the number in units place is same as digit in units place of number 3^3. Hence answer is 7.

edited Nov 05 '17 at 17:50

answered Nov 05 '17 at 17:43

score 2 · Answer 3 · answered Nov 05 '17 at 17:38

2

We observe $3^{2011}\mod 10$ to get the last digit.

It is $3^{2011}=3\cdot 3^{2010}=3\cdot 9^{1005}\equiv 3\cdot (-1)^{1005}\equiv -3\equiv 7\mod 10$

answered Nov 05 '17 at 17:38

Cornman

11,065
4
30
57

score 2 · Answer 4 · answered Nov 05 '17 at 17:38

The last digits of 3 repeat every 4 powers: $3^1=3$, $3^2=9$, $3^3=27$, $3^4=81$, and because the last digit of $3^4$ is one, multiplying by three just gives us three again, and the pattern continues.

So taking $2011 \mod 4$, we get $3 \mod 4$, so the last digit is the same as that of $3^3=27$, which is 7.

score 2 · Answer 5 · answered Nov 05 '17 at 17:40

$3^1$: 3 is last digit

$3^2$: 9 is last digit

$3^3$: 7 is last digit

$3^4$: 1 is last digit

$3^5$: 3 is last digit

$3^6$: 9 is last digit

$3^7$: 7 is last digit

$3^8$: 1 is last digit

$(3,9,7,1)$ is period.

$$3^{2011}=3^{502\cdot4+3}$$ Then 7 is last digit

Guy Fsone · Answer 6 · 2017-11-06T10:55:12.720

2

the last digit correspond to $$3^{2011}\equiv x\mod 10$$ We have, $$3^4\equiv 1\mod 10$$

But, $$2011 = 4 * 502+3$$

Thus, $$3^{2011}=3^{4*502+3} \equiv 3^3\mod 10~~~\text{ that is }~~ 3^{2011} \equiv 7\mod 10 $$

Hence $x=7 $ is the last digit

edited Nov 06 '17 at 10:55

answered Nov 05 '17 at 17:40

Guy Fsone

23,903

What is the last digit of $3^{2011}$

6 Answers6