Htykuut

$13^4$ 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.

Using the fact that $a^4equiv1$ mod $5$ if $5notmid a$, we find

$$11^{11}+12^{12}+13^{13}equiv
begin{cases}1+0+1equiv0mod 2\
1+1+3equiv0mod 5
end{cases}$$

Added later: Here is an alternative way to show that $11^{11}+12^{12}+13^{13}$ is a multiple of $10$ without resorting to modular arguments. All that's needed is the algebraic identity $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+cdots+xy^{n-2}+y^{n-1})$ with various interpretations of $x$, $y$, and $n$ (e.g., $x=12^4$, $y=6^4$ and $n=3$).

Note that $1+6+13=20$ is a multiple of $10$. It follows that $11^{11}+12^{12}+13^{13}$ is a multiple of $10$ if and only if $(11^{11}-1)+(12^{12}-6^{12})+(6^{12}-6)+(13^{13}-13)$ is a multiple of $10$. But

$$begin{align}
11^{11}-1&=(11-1)(11^{10}+11^9+cdots+11+1)\
&=10(11^{10}+11^9+cdots+11+1)\
12^{12}-6^{12}&=(12^4-6^4)(12^8+12^46^4+6^8)\
&=(12^2+6^2)(12^2-6^2)(12^8+12^46^4+6^8)\
&=(144+36)(12^2-6^2)(12^8+12^46^4+6^8)\
&=180(12^2-6^2)(12^8+12^46^4+6^8)\
6^{12}-6&=6(6^{11}-1)\
&=6(6-1)(6^{10}+6^9+cdots+1)\
&=30(6^{10}+6^9+cdots+1)\
13^{13}-13&=13(13^{12}-1)\
&=13(13^4-1)(13^8+13^4+1)\
&=13(13^2+1)(13^2-1)(13^8+13^4+1)\
&=13(169+1)(13^2-1)(13^8+13^4+1)\
&=13cdot170(13^2-1)(13^8+13^4+1)\
end{align}$$

Remark: This approach is, of course, more convoluted than the modular approach. (It also relies in part on knowing which terms, such as $12^2+6^2$, will produce a factor of $10$.) If anything, it should demonstrate the value of learning how modular arithmetic works!

You might be able to show that the units digit is zero pretty easily.

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.

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
$11^{11} + 12^{12} + 13^{13} equiv 1+6+3 equiv 0 (mod 10)$

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 $n^4equiv 1$ and that applies with $n=3$ so that $13^{13}equiv 3^{12}cdot 3equiv 3 bmod 10$.

$12$ 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 $m^2equiv m$ 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.

You can easily prove it without any modular arithmetic. Just look at the last digits.

$$3^1 = color{blue}{3} quad 3^2 = color{blue}{9} quad 3^3 = 2color{blue}{7} quad 3^4 = 8color{blue}{1} implies 3, 9, 7, 1, 3, 9, 7, 1, ...$$

$$2^1 = color{green}{2} quad 2^2 = color{green}{4} quad 2^3 = color{green}{8} quad 2^4 = 1color{green}{6} implies 2, 4, 8, 6, 2, 4, 8, 6, ...$$

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.

$$13 = color{purple}{3}(4)+color{blue}{1} quadtext{Three loops done; first exponent}$$

$$12 = color{purple}{2}(4)+color{green}{4} quadtext{Two loops done; fourth exponent}$$

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?

Tedious the first time you do it way but once you do it is obvious:

$$11^{11} = (10 + 1)^{11} = 10^{11} + 11*10^{10} + 55*10^9..... + 55*10^2 + 11*10 + 1 = 10M + 1$$ for $M= 10^{10} + 11*10^{9} + 55*10^8..... + 55*10 + 11$. But ... THE EXACT VALUE OF $M$ WILL NOT MATTER!!!... The only thing that matters is the remander $1$.

Likewise $$12^{12} = (10 + 2)^{12} = 10^{12} + 12*2*10^{11} + 66*2^2*10^{10} + .... . + 66*2^{10}*10^2+ 12*2^{11}*10 + 2^{12} = 10N + 2^{12}$$ for $N = 10^{12} + .... text{oh, what that hell do we care}$.

And $$13^{13} = (10+3)^{13} = 10^{13} + 13*10^{12}*3 + ..... +13*10*3^{12} + 3^{12} = 10K + 3^{13}$$ for $K = text{whatever}$.

So $$11^{11}+12^{12}+13^{13} = 10N + 10M + 10K + 1 + 2^{12} + 3^{13}$$. But since we are only interested in whether it is divisible by $10$ we can ignore the $10N + 10M + 10K$ because it is divisible by $10$.

So for all that tedious work we've come up with the simple and useful idea that "if we want to find out if something is divisible by $10$ we only have to do math on the remainders. We could have done that in our heads!

If $equiv_{10}$ means "has the same remainder when divided by $10$" we could have done that in one line:

$$11^{11} + 12^{12} + 13^{13}equiv_{10} 1^{11} + 2^{12} + 3^{13}$$

$1^{11} = 1$ of course.

And we can do $2^{12}$ in steps.

$2^2 = 4; 2^4= 4^2 = 16=10+6equiv_{10} 6$.

$2^8 = (10 + 6)^2 = 100 + 2*6*10 + 36= 10M + 6equiv{10} 6$.

So $2^{12} = 2^8*2^4 = (10M + 6)(10 +6) =100M + 6*10 + 10*6M + 6*6equiv_{10}6*6=36equiv_{10} 6$.

At this point we should realize we can just go deirectly to working with remainders.

$3^{2} = 9; 3^4 = 9^2 = 81 equiv_{10} 1$. So $3^{13}3^4*3^4*3^4*3equiv_{10} 1*1*1*3=3$.

So if we had used this concept of remainders from the begining we'd have done it in two or three lines:

$11^{11} + 12^{12} + 13^{13} equiv_{10}$

$1^{11} + 2^{12} + 3^{13} equiv_{10}$

$1 + (2^4)^3 +(3^4)^3 times 3 equiv_{10}$

$1 + 16^3 + 81^3*3 equiv $

$1 +6^3 + 1^3*3equiv $

$1 + 36*6 + 3 equiv $

$1 + 6*6 + 3equiv $

$1 + 36 + 3equiv $

$1 + 6 + 3equiv 10 equiv 0$.

So $11^{11} + 12^{12} + 13^{13}$ has remainder $0$ when divided by $10$ which is to say $11^11 + 12^{12} +13^{13}$ is divisible by $10$.

====

A number is devisible by $10$ if and only if the last digit is $0$.

And value of the last digits arithmetically "distribute". That is if the last digit of $n$ is $a$ then the last digit of $n^k$ is the last digit of $a^k$. (Because $n = 10b +a$ so $n^k =(10b + a)^k = abunchofmultiplesof(10) + a^k$) and if the last digit of $n$ is $a$ and the last digit of $m$ is $c$ then the last digit of $m + n$ is the last digit of $b+1$ and the last digit of $mtimes n$ is the last digit of $btimes c$.

So $11^{11}+12^{12}+13^{13}$ is divisible by $10$ if only if its last digit is $0$.

And the last digit of $11^{11}+12^{12}+13^{13}$ is the last digit of $1^{11}+2^{12}+3^{13}$

$1^{11} = 1$.

$2^{12} = 2^4times 2^4 times 2^4 = 16times 16times 16$ so the last digit is the same as $6times 6times 6$ so the last digit is the same as $6times 6=36$ so the last digit is $6$.

$3^{13} = 3^4times 3^4 times 3^4 times 3$. $3^4 = 81$ so the last digit of $3^{13}$ is the last digit of $1times1times 1times 3 = 3$>

So the last digit of $11^{11}+12^{12}+13^{13}$ is the same as the last digit of $1 + 6 + 3 =10$. And that digit is $0$.

So, yes, $11^{11}+12^{12}+13^{13}$ is divisible by $10$.

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 $,a,b,cinBbb Z, k,nin Bbb N$

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 $ rmcolor{#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 } rmcolor{#0a0}{BT} as above\[.2em]
&= (1!+!15)^{large n}! + 5,i\[,2em]
&= 1 + 5,j, +, 5,i, text{for an integer } j, text{by } rmcolor{#0a0}{BT} as above\
end{align}$

Similarly $, (1+10a)^{large k} =, 1 + 5,d$

and also $ (3+10c)^{large 1+4n}! = 3 + 5,e$

Therefore their sum equals $: 1 + 1 + 3 + 5(cdots) = $ multiple of $5. $ QED

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 $10$

Note that $$11equiv1;mathtt{mod}(10) Rightarrow {11^{11}}equiv {1^{11}}equiv1 ;mathtt{mod}(10)$$
Analugously $${12^{12}}equiv{2^{12}}equiv{(2^6)^2}equiv{64^2}equiv6;mathtt{mod}(10)$$
$${13^{13}}equiv{(13^4)^2}*13equiv{1^2}*3equiv3; mathtt{mod}(10)$$
Thus
$${11^{11}}+{12^{12}}+{13^{13}}equiv1+6+3equiv 0; mathtt{mod}(10)$$
$$therefore10mid bigl({11^{11}}+{12^{12}}+{13^{13}}bigr)$$

You can alternatively use the binomial theorem:
$$11^{11}=(10+1)^{11}=sum_{k=0}^{11}binom{11}{k}10^k=binom{11}{0}10^0+binom{11}{1}10^1+binom{11}{2}10^2+ldots+binom{11}{11}10^{11}$$
$$=1+11*10+55*100+ldots+10^{11}=1+10Bigl(11+550+ldots+10^{10}Bigr)$$

$$12^{12}=(10+2)^{12}=sum_{k=0}^{12}binom{12}{k}10^k2^{12-k}=binom{12}{0}10^02^{12}+binom{12}{1}10^12^{11}+binom{12}{2}10^22^{10}+ldots+binom{12}{12}10^{12}=2^{12}+12*10*2^{11}+66*100*2^{10}+ldots+10^{12}$$ $$=2^{12}+10Bigl(12*2^{11}+660*2^{10}+ldots+10^{11}Bigr)$$

$$13^{13}=(10+3)^{13}=sum_{k=0}^{13}binom{13}{k}10^k3^{13-k}=binom{13}{0}10^03^{13}+binom{13}{1}10^13^{12}+binom{13}{2}10^23^{11}+ldots+binom{13}{13}10^{13}=3^{13}+13*10*3^{12}+78*100*3^{11}+ldots+10^{13}$$ $$=3^{13}+10Bigl(13*3^{12}+780*3^{11}+ldots+10^{12}Bigr)$$

Thus $${11^{11}}+{12^{12}}+{13^{13}}$$ $$=1+10Bigl(11+550+ldots+10^{10}Bigr)+2^{12}+10Bigl(12*2^{11}+660*2^{10}+ldots+10^{11}Bigr)+3^{13}+10Bigl(13*3^{12}+780*3^{11}+ldots+10^{12}Bigr)$$ $$=1+2^{12}+3^{13}+10Biggl(cdotsBiggr)$$

Now $$1+2^{12}+3^{13}=1+2^{3*4}+3^{3*4}*3=1+8^4+9^4*3=1+64^2+81^2*3$$ $$=1+4096+6561*3=23780Rightarrow 10mid (1+2^{12}+3^{13})$$

And we would be done.

PS:

I still find the modular approach easier and more elegant...

Assuming you have knowledge of modular arithmetic.

$11equiv 1mod 10implies 11^{11}equiv 1mod 10$.

$12equiv 2mod 10implies 12^6equiv 2^6 mod10equiv 4mod 10$ $implies 12^{12}equiv 4^2mod 10equiv 6mod 10$.

$13equiv 3 mod 10implies 13^4equiv 3^4 mod 10equiv 1 mod 10implies 13^{13}equiv 3mod 10.$

So $11^{11}+12^{12}+13^{13}equiv 1+6+3 mod 10equiv 0 mod 10.$

Hence $11^{11}+12^{12}+13^{13}$ is divisible by 10.

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 $mathbb{Z}/10mathbb{Z}$ so $3^{13}=3^{12}*3=3$.