# The last digit of $2^{2006}$

My $$13$$ year old son was asked this question in a maths challenge. He correctly guessed $$4$$ on the assumption that the answer was likely to be the last digit of $$2^6$$. However is there a better explanation I can give him?

• Because last digit of $2^5$ is 2, so last digit of $2^{2006}$ is last digit of $2\cdot 2^{401}$, and so on. Jan 23, 2013 at 9:25
• Note that $2^4\equiv 6 \bmod 10$ , $6^n\equiv 6 \bmod 10$ and the last digit is the no. mod 10. So $2^n\equiv 2^x\bmod 10$ if $x\equiv n \bmod 4$ 2006 mod 4=2 so the last digit is that of 4 Jan 23, 2013 at 9:32
• I remember this exact problem from when I had math competitions many, many years ago... Jan 24, 2013 at 2:28
• Let's say the last digit of $2^{2006} = x$ then $x \equiv 2^{2006} (mod 10)$, to solve this "reduce" the exponent checking or exponentiating 2 to some power that is a divisor of 2006 and also that you know and isn't too big. For example: $2^{2006}=4^{1003}=16^{500}4^3$ etc... Keep in mind that $16^{500}4^3 \equiv 6^{500}4^3 (mod 10)$, and also remember that you can't apply in this case Fermat-Euler's theorem because $(10,2)=2$. Sep 28, 2014 at 0:42
• @Ishan Banarjee can you elaborate your comment on the solution? Dec 31, 2020 at 15:45

$$2^{4} = 16$$. Multiply any even integer by $$6$$ and you don't change the last digit: $$0 \times 6 = 0$$, $$2 \times 6 = 12$$, $$4 \times 6 = 24$$ etc. The same is true if you multiply an even integer by anything whose last digit is $$6$$, in particular by $$16$$. Now $$2006 = 2004 + 2$$ where $$2004 = 501 \times 4$$, so $$2^{2006} = (2^4)^{501} \times 2^2$$ has the same last digit as $$2^2$$.

• Many thanks Robert Jan 23, 2013 at 9:49
• I don't find this particularly satisfying. "Multiply any even integer by 6 and you don't change the last digit." But how does that explain anything to the 13 year old? To them, that's just a bit of unexplained magic that they (probably) didn't know beforehand. Jan 23, 2013 at 21:43
• For a $13$-year-old, is it better to say that $6 \equiv 1 \mod 5$ and $\equiv 0 \mod 2$ ...? I doubt it. Jan 23, 2013 at 21:54
• Multiple of 6: Assumption x * 6 = 10*u + x whereas x in {0,2,4,6,8} and u is unknown. Let's say x = 2*y for y in {0..4}. Then 2*y*6 = 2*y*5 + 2*y = 10*y + x.
– user59502
Jan 24, 2013 at 13:16
• Or more simple: x*6 = 5*x + x. Since x is even, 5*x contains at least one 10 in it's integer factorization and therefore 5*x won't affect the last digit. "+x" will be the only part which affects the last digit.
– user59502
Jan 31, 2013 at 7:19

Well, looking at successive powers of two, starting at $2^1$ we have $$2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, \dots$$

Now, looking at only the final digit we have $$2, 4, 8, 6, 2, 4, 8, 6, 2, 4, \dots$$

You may be able to guess that the pattern repeats the segment $2, 4, 8, 6$ forever. If you're correct, then you can work out the last digit by figuring where $2^{2006}$ sits in this pattern. Note that the first term in the sequence is $2$, and every four terms along it is also $2$. That is the $(4k+1)^{\textrm{th}}$ term is always $2$. Likewise, the $(4k + 2)^{\textrm{nd}}$ term is always $4$, the $(4k+3)^{\textrm{rd}}$ term is always $8$, and the $(4k+4)^{\textrm{th}}$ term is always $6$. Now you just need to figure out whether $2006$ is of the form $4k + 1, 4k+2, 4k + 3$, or $4k + 4$. As $2006$ is even, it is either of the form $4k+2$ or $4k+4$. As $4k+2 = 2006$ has an integer solution $(k = 501)$ and $4k + 4 = 2006$ doesn't, $2006$ is of the form $4k+2$ and therefore $2^{2006}$ ends in a $4$.

A guess is all well and good, but it ain't no proof. How do we know the pattern of digits continues to repeat forever (or at least up to $2006$)? Well, you can use modular arithmetic to prove the pattern continues, but if your son knew modular arithmetic, he could have used it to solve the problem in the first place.

Another way to see the pattern continues is to think about what happens when you multiply numbers. Consider multiplying the number $43$ by $2$. How do you do it? First you multiply $3$ and $2$ and put the result in the ones column, then you move onto the $4$ and multiply it by $2$ and put it in the tens column. Had we chosen $46$ instead of $43$ then it is slightly more complicated because $6\times 2 = 12$. In this situation we put the $2$ in the ones column and we carry the $1$ to the tens column. This all sounds a little bit boring but the point is this:

To find out the last digit of $a\times 2$, where $a$ could have lots of digits, we just need to know what (the last digit of $a$)$\, \times\, 2$ is. In particular, the last digit of $($(the last digit of $a$)$\, \times\, 2)$ is the same as the last digit of $(a\times 2)$.

This is true because of the way we multiply as explained above. When multiplying $43$ by $2$, we only put something in the ones column when we did $3\times 2$. In the case of $46$ multiplied by $2$, we get $6\times 2 = 12$ and put the $2$ in the ones column. So irrespective of the size of the number, we only put something in the ones column when we multiply the rightmost (i.e. last) digit by $2$, and in the case that the result has more than one digit, we only take the last digit of the result and put it in the ones column.

With this in mind, we have the final digits of the first four powers of $2$ are $2, 4, 8, 6$. As $6 \times 2 = 12$, we know that the final digit of $2^5 = 2^4\times 2$ is $2$. Then as $2\times 2$ we know that the final digit of $2^6 = 2^5\times 2$ is $4$. Then as $4\times 2= 8$ we know the final digit of $2^7 = 2^6\times 2$ is $8$. Then as $8\times 2 = 6$, we know that the final digit of $2^8=2^7\times 2$ is $6$. Now we're back to $6$ and can see that we will continue the pattern forever. So the initial guess was correct and hence the argument that $2^{2006}$ ends in a $4$ is valid.

Note, similar considerations to those made above can be used to show that the last digit of $a\times b$ is the same as the last digit of (last digit of $a$)$\, \times\,$(last digit of $b$). As before, you can also prove this using modular arithmetic.

• This is the answer I would have given. Jan 23, 2013 at 13:58

The number may be written as $$2^{2006}=4^{1003}$$ $4^1=4$
$4^2=4\times4=16$
$4^3=4\times4\times4=64$
$4^4=4\times4\times4\times4=256$
At this point we immediately see that if the power of $4$ is odd then the last digit is $4$ otherwise is $6$.

• How does this have 46 upvotes? This is a really common method?? Oct 23, 2014 at 11:04
• @Katie Yeah, it is a common method for me. Oct 23, 2014 at 11:19
• Being a normal or common method doesn't mean one cannot like it. @Katie Jan 26, 2016 at 11:55
• @Jasser You believe this answer warrants 530 reputation points? About 46 of your answers worth of reputation. Feb 1, 2016 at 9:06
• Well it certainly doesnt but people follow people I.e., when they see some answer with many upvotes they tend to upvote it.... and moreover most of the people will try to answer the question with $2$ instead of $4$.... and so they might have liked it since they could not think of it and I am one of them..... Also the question doesnt have to have so many upvotes and so do other answers.... but it just happens as I said..... Feb 1, 2016 at 15:35

If $a\equiv b\pmod n\implies a^m\equiv b^m\pmod n$

Thus, $2^5\equiv 2\pmod {10}\implies (2^5)^{400}\equiv 2^{400}\equiv2^{80}\equiv 2^{16}\equiv2^3.2\equiv 6\pmod{10}$

Therefore, $2^{2006}\equiv 2^{2000}.2^6\equiv6.4\pmod{10}\equiv 4\pmod{10}$

Hence, the unit digit is $4$

• I see this to be better & correct explanation than others Jan 23, 2013 at 14:56
• This is the best indeed Jan 23, 2013 at 15:10

The last digit of $2^{2006}$ is $2^{2006} (\text{mod }10)$. So let's look at the behavior of powers of $2$ mod $10$.

$2^2 = 4$, $2^3 = 8$ (nothing interesting so far), $2^4 = 16 = 6$ (remember we are working mod $10$, we only keep the last digit). $2^5 = 32 = 2$, Now THIS is interesting. This suggests we divide $2006$ by $5$.

So let's do that: $2006 = (401)(5) + 1$

going back to working mod $10$, we have:

$2^{2006} = 2^{(5)(401)+1} = (2^5)^{401}(2)= 2^{402}$ so we have knocked down the size of our exponent a great deal. Repeating this procedure again leads to:

$2^{402} = 2^{(5)(80) + 2} = (2^5)^{80}(2^2) = 2^{82}$ (still mod $10$).

Another go:

$2^{82} = 2^{(5)(16)+2} = (2^5)^{16}(2^2) = 2^{18}$ (mod $10$),

One last time:

$2^{18} = 2^{(5)(3)+3} = (2^5)^3(2^3) = 2^6$ We can stop now, it's clear that the last digit we are looking for is $4$ (the last digit of $64 = 2^6$). But if we wished (and we were for some reason unwilling to physically compute $2^6$) we could continue for one last step:

$2^6 = (2^5)(2^1) = (2)(2) = 4$ (mod $10$).

When doing modular exponentiation, Euler's Totient function ($\phi$) is quite handy.

To get the last digit of $2^{2006}$, we simply rewrite it as $x \equiv 2^{2006}\; (mod\;10)$.

For any modular exponentiation, we can express $a^b\;(mod\;c)$ more simply as $a^{b (mod\;\phi(c))}\;(mod\;c)$.

$\phi(10)=4$, therefore $2^{2006}\equiv2^{2006\:(mod\:4)}\equiv2^2\equiv4\;(mod\; 10)$

• While a very useful tool, using Euler's Theorem here sort of trivializes the problem. Jan 23, 2013 at 22:17
• Are you sure this is right? Sure, the answer is, but if I recall correctly Euler's theorem requires $\gcd(a,m)=1$ and says in that case $a^{\phi(m)}\equiv 1\pmod{m}$, but here we have $\gcd(2,10)=2\neq1$. Or am I missing something? The chinese remainder theorem could be applied here though: $2^{2006}\equiv 0 \pmod{2}$ and $2^{2006}\equiv 4 \pmod{5}$ and then apply Euclid's extended algorithm or just notice immediately that $4\equiv 0\pmod{2}$ and $4\equiv 4\pmod{5}$ (in the preceding steps I have applied Euler's theorem which was possible because 5 and 2 are prime.)
– user50407
Jan 24, 2013 at 14:04

$4$ because $2\times 2\times\dots\times 2$ ($2006$ times); if number in power has remainder $0$, $1$, $2$, or $3$ when divided by $4$, then last digit is $6$, $2$, $4$ or $8$ respectively.

• I have slightly reworded your answer. If it is not what you meant and you want to change it back, please do so. Jan 24, 2013 at 12:08

The last digit of the number is the remainder of division by 10, since any number is represented as: $$a_n \cdot 10^n + a_{n-1} \cdot 10^{n-1} + \ldots + a_1 \cdot 10 + a_0 = \overline{a_na_{n-1}\ldots a_1a_0}$$ $$\quad \Leftrightarrow \quad$$ $$a_n \cdot 10^n + a_{n-1} \cdot 10^{n-1} + \ldots + a_1 \cdot 10 + a_0 \equiv a_0 \mod 10.$$ Therefore, it is enough to consider the remainder of division by $$10$$ of the original number.

According to the Chinese remainder theorem $$2^{2006} \equiv x \mod 10 \quad \Leftrightarrow \quad \begin{cases} 2^{2006} \equiv x \mod 5 \\ 2^{2006} \equiv x \mod 2 \end{cases} \quad \Leftrightarrow \quad \begin{cases} 2^{2006} \equiv x \mod 5 \\ 0 \equiv x \mod 2\end{cases}$$

Since the Euler function of 5 is equal to $$\phi(5) = 5-1 = 4$$ (as well as for any prime number), then by Euler's theorem the system of comparisons can be rewritten as:

$$\begin{cases} 2^2 \equiv x \equiv 4 \mod 5 \\ 0 \equiv x \mod 2\end{cases}$$

Going through all the numbers from $$0$$ to $$10$$, we make sure that only $$4$$ is suitable, so $$x = 4$$, which will be the last digit of the number.