What is the last digit of $2003^{2003}$?

Question

What is the last digit of this number?

$$2003^{2003}$$

Thanks in advance.

I don't have any kind of idea how to solve this. Except that $3^3$ is $27$.

Answer 1

When calculating the last digit of $a\cdot b$ you only need to find the product of the last digit of $a$ and the last digit of $b$.

so $2003^{2003}$ ends in the same digit as $3^{2003}$

$3^1$ ends in $3$

$3^2$ ends in $9$

$3^3$ ends in $7$

$3^4$ ends in $1$

$3^5$ ends in $3$

$3^6$ ends in $9$

$3^7$ ends in $7$

$3^8$ ends in $1$.

So $3^{4k}$ ends in $1$.

From here $3^{2000}$ ends in $1$ (since $2000$ is $4\cdot 500$).

Therefore $3^{2001}$ ends in $3$, $2^{2002}$ ends in $9$ and $3^{2003}$ ends in $7$

Answer 2

Let's look at the residue mod $10$. $2003^{2003} \equiv 3^{2003} \equiv 3^{2000} \times 3^{3} \equiv 3^{4 \times 500} \times 3^{3} \equiv 7$. I hope I'm not mistaken!

Answer 3

Since we have

$$3^1=\color{red}{3},3^2=\color{red}{9},3^3=2\color{red}{7},3^4=8\color{red}{1},3^5=24\color{red}{3},\cdots.$$

and $$2003\equiv 3\pmod 4,$$ the right-most digit of $2003^{2003}$ is the same as the right-most digit of $3^3=27$.

Answer 4

Notice that $2003 = 2000 + 3$, and that when doing the binomial expansion of $(2000 + 3)^{2003}$ you will get many terms, but what will determine the last digit will be the $3^{2003}$ because all of the others end in $0$.

Now, the remainders when dividing by 3 after exponentiating it to some power are periodic.

$3^1 = 3$

$3^2 = 9$

$3^3=27$

$3^4=81$

$3^5=273$

Now, $3^5 \equiv 3^1 (mod 10) $, and the period turns out to be 4.

So, when finding the last digit, we only need to know the remainder of the exponent divided by 4. Notice that that remainder is 3. Therefore, $3^{2003} \equiv 3^3 (mod 10)$, which is just the remainder of $27/10$, which is 7. That's the last digit you're looking for.

Answer 5

$$2003^{2003} = (200*10+3)^{2003}$$ Throwing out every term that doesn't influence the last digit $$=3^{2003} = 3^3 \cdot (3^4)^{500} = 3^3 \cdot (8*10+1)^{500}$$ Again throwing out every term that doesn't influence the last digit $$=3^3 \cdot (1)^{500}= 3^3 = 2*10+7 $$ Last time throwing out every term that doesn't influence the last digit $$= 7$$

Answer 6

$$\begin{align} 2003^{2003} &= 3^{2003} &\pmod{10} \\ &= 9^{1001} \times 3 &\pmod {10} \\ &= (-1)^{1001} \times 3 &\pmod {10} \\ &= -1 \times 3 &\pmod {10} \\ &= 7 &\pmod {10} \end{align}$$

Answer 7

When we find "the last digit" of a number, it means we are working "mod $10$" (the same as "regular" arithmetic, but we only pay attention to the last digit). In particular, $ab$ (mod $10$) is the same as $[(a$ (mod $10))\cdot (b$ (mod $10))]$(mod $10$).

For example, the last digit of $37\cdot 22 = 814$ is the same as $7\cdot 2= 14$, namely: $4$.

It turns out that $3^4 = 81$ has a last digit of $1$, so:

$2003^{2003} = 3^{2003}$ (mod $10$) = $3^{750\cdot 4 + 3}$ (mod $10$)

$=(3^4)^{750}(3^3)$ (mod $10$) = $1^{750}(3^3)$ (mod $10$) = $3^3$ (mod $10$).

I daresay the last digit of $3^3$ is easily discoverable.

It isn't immediately apparent that for any number $a$ with last digit $k\in U = \{1,3,7,9\}$, that the last digit of $a^4$ is $1$, so that to find the last digit of $a^m$ we only need to find $k^{m'}$ (where $m'$ is the remainder of $m$ upon division by $4$) but this is a consequence of a far-reaching number-theoretical result known as Euler's Theorem. The down-side to using this, is that problems easy to state (such as "what is the last digit of (insert very large number here)") can be hard to explain the solutions of, without recourse to more mathematics than it looks like would be involved.