Show $11^{11}+12^{12}+13^{13} =10k$ without direct calculation Prove that $11^{11}+12^{12}+13^{13}$ is divisible by $10$.
Obviously you could just put that in to a calculator and see the results, but I was wondering about some of the other approaches to this? I have not studied modulus', so if you could explain it without them, it would be better for me. Thanks!
 A: $13^4$ ends with $1$ so does $13^{12}$ and thus $13^{13}$ ends with $3$
$12^4$ ends with $6$ so $12^{12}$ also ends with $6$
$11^{11}$ ends with $1$ so your statement is valid.
A: You might be able to show that the units digit is zero pretty easily.
A: Last digit of $11^{11}$ is obviously 1.
$12^1$ ends at 2. $12^2$ ends at 4. $12^3$ at 8. $12^4$ at 6, and $12^5$ again at 2. So, last digit of $12^{12}$ is 6.
$13^1$ ends at 3. $13^2$ at 9. $13^3$ at 7, $13^4$ ends at 1 and $13^5$ again ends at 3. Hence, $13^{13}$ ends at 3.
Finally, $11^{11}+12^{12}+13^{13}$ ends at 0 at your number is divisible by 10.
A: Simple calculation :
$11 \equiv 1 (mod 10)$ $\implies$ $11^{11} \equiv 1 (mod 10)$
$12 \equiv 2 (mod 10)$ $\implies$ $12^{12} \equiv 6 (mod 10)$
$13 \equiv 3 (mod 10)$ $\implies$ $13^{13} \equiv 3 (mod 10)$
Hence
$11^{11} + 12^{12} + 13^{13} \equiv 1+6+3 \equiv 0 (mod 10)$
A: The final digit of $11^{11}$ is easy. If we are able to use Fermat-Euler we have that $\varphi (10)=4$ so that if $n$ is coprime to $10$ then $n^4\equiv 1$ and that applies with $n=3$ so that $13^{13}\equiv 3^{12}\cdot 3\equiv 3 \bmod 10$.
$12$ is more interesting - we can't use the coprime property. The exponent is small enough that you can do it by hand, but here is another trick (working modulo $10$).
First $12^{12}\equiv 2^{12}$
Then $12^{12}=3^{12}\cdot 4^{12}\equiv \left(2^{12}\right)^2$ (we did $3^{12}\equiv 1$ already)
Let $2^{12}\equiv m$, then $m^2\equiv m$ from the two calculations above, and $m$ is even and non-zero, so must be $6$.

I added this because it is interesting, and very occasionally such alternative ways of working save a lot of time.
A: You can easily prove it without any modular arithmetic. Just look at the last digits.
$$3^1 = \color{blue}{3} \quad 3^2 = \color{blue}{9} \quad 3^3 = 2\color{blue}{7} \quad 3^4 = 8\color{blue}{1} \implies 3, 9, 7, 1, 3, 9, 7, 1, ...$$
$$2^1 = \color{green}{2} \quad 2^2 = \color{green}{4} \quad 2^3 = \color{green}{8} \quad 2^4 = 1\color{green}{6} \implies 2, 4, 8, 6, 2, 4, 8, 6, ...$$
The pattern loops every $4^{th}$ exponent, as you can see. Notice that $13^{13}$ must end with whatever $3^{13}$ ends with, and $2^{12}$ must end with whatever $2^{12}$ ends with.
$$13 = \color{purple}{3}(4)+\color{blue}{1} \quad\text{Three loops done; first exponent}$$
$$12 = \color{purple}{2}(4)+\color{green}{4} \quad\text{Two loops done; fourth exponent}$$
Thus, $13^{13}$ ends with $\color{blue}{3}$ and $12^{12}$ ends with $\color{green}{6}$. $11^{11}$ obviously ends with $1$, so what does the unit digit become?
A: Since you don't know modular arithmetic we can instead employ the Binomial Theorem.
Theorem $\ 10\ $ divides $\ (1\!+\!10a)^{\large k}\!+ (2\!+\!10b)^{\large  4n}\!+(3\!+\!10c)^{\large 1+4n}\ $ for $\,a,b,c\in\Bbb Z,\ k,n\in \Bbb N$
Proof $\ $ To show it's divisible by $10$ it suffices to show it's divisible by $\,2\,$ and $\,5\,$. Notice it has parity odd + even + odd = even, so $\,2\,$ divides it. Below we show that $\,5\,$ divides it too.
Note that $\ \ (x\,+\,\color{#c00}5y)^{\large m}\, =\,\ x^{\large m}\, +\, \color{#c00}5(\cdots)\ \ \ \,$ by $\ \rm\color{#0a0}{BT}$ := Binomial Theorem, for $(\cdots)$  an integer.
$\!\!\begin{align} {\rm Therefore}\, \ 
(2+10b)^{\large 4n} &=\  2^{\large 4n}\,+\, 5\,i,\ \ \ \ \ \ \ \ \, \text{for an integer } i, \ \text{by } \rm\color{#0a0}{BT}\ as\ above\\[.2em]
&= (1\!+\!15)^{\large n}\! + 5\,i\\[,2em]
&=\  \color{darkorange}1 + 5\,j\, +\, 5\,i, \ \ \ \text{for an integer } j, \ \text{by } \rm\color{#0a0}{BT}\ as\ above\\
\end{align}$
Similarly $\,\ (1+10a)^{\large k}\ \ =\,\ \color{#0a0}1 + 5\,d$
and also $\ \ (3+10c)^{\large 1+4n}\! = \color{#90f}3 + 5\,e$
Therefore their sum equals $\: \color{darkorange}1 + \color{#0a0}1 + \color{#90f}3 + 5(\cdots) = $ multiple of $5.\ \ $ QED
A: The first digit of $11^{11}=$ the first digit of $1^{11}=$ $1$.

The first digit of $12^5$ equals the first digit of $2^5=$ the first digit of $32=2$.
$\text{(The first digit of $12^{10}) =$ (The first digit of $12^5)^2 = 4$ }$.
$\text{(The first digit of $12^{12}) =
       $ (The first digit of $12^{10}) \times($The first digit of $12^2)  = 6$ }$.

$\text{(The first digit of $13^{4}) =$ (The first digit of $3^4) = 1$ }$.
$\text{(The first digit of $13^{12}) =$ (The first digit of $(13^4)^3) = 1$ }$.
$\text{(The first digit of $13^{13}) =$ (The first digit of $13^{12} \times 3) = 3$ }$.

$\text{(The first digit of $11^{11} + 12^{12} + 13^{13}) =$
       (The first digit of $1+6+3) = 0$ }$.

Hence $11^{11} + 12^{12} + 13^{13}$ is a multiple of $10$
A: It is well known that
$\tag 1 10 \displaystyle \text{ divides } a + 10 \cdot b \; \text{ if and only if }\; 10 \text{ divides } a $
We'll now construct a list of numeric expressions separated by $//$; either all the expressions are divisible by $10$ or none of them are. The OP should be able to follow the logic, which uses $\text{(1)}$ over and over, sometimes using it more than once in one step.
$\displaystyle 11^{11}+12^{12}+13^{13} \quad // $
$\displaystyle (10+1)\cdot 11^{10}+(10+2)\cdot12^{11}+(10+3)\cdot13^{12} \quad // $
$\displaystyle 1^{10} \cdot 11 +2^{11}\cdot 12+3^{12} \cdot 13 \quad // $
$\displaystyle 1 +2^{12}+3^{13}  \quad // $
$\displaystyle 1+4^{6}+3\cdot3^{12} \quad // $
$\displaystyle 1+16^{3}+3\cdot9^{6} \quad // $
$\displaystyle 1+6^{3}+3\cdot81^{3} \quad // $
$\displaystyle 1+6\cdot36+3\cdot 1^{3} \quad // $
$\displaystyle 1+6\cdot6+3 \quad // $
$\displaystyle 40$
A: Assuming you have knowledge of modular arithmetic.
$11\equiv 1\mod 10\implies 11^{11}\equiv 1\mod 10$.
$12\equiv 2\mod 10\implies 12^6\equiv 2^6 \mod10\equiv 4\mod 10$ $\implies 12^{12}\equiv 4^2\mod 10\equiv 6\mod 10$. 
$13\equiv 3 \mod 10\implies 13^4\equiv 3^4 \mod 10\equiv 1 \mod 10\implies 13^{13}\equiv 3\mod 10.$
So $11^{11}+12^{12}+13^{13}\equiv 1+6+3 \mod 10\equiv 0 \mod 10.$
Hence $11^{11}+12^{12}+13^{13}$ is divisible by 10.
A: When calculating in (mod 10) we get $11^{11}+12^{12}+12^{13}=1^{11}+2^{12}+3^{13}=1+6+3=10$. Here I knew $2^{10} =1024$, so $2^{12}=6(mod10)$ and $3$ has order $4$ in $\mathbb{Z}/10\mathbb{Z}$ so $3^{13}=3^{12}*3=3$.
