Show $11^{11}+12^{12}+13^{13} =10k$ without direct calculation

Question

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!

Answer 1

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

Answer 2

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.

Answer 3

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 $3^{13}=3^{12}*3=3$.

Answer 4

$13^4$ 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.

Answer 5

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.

Answer 6

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.

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

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

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

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 $3^{13}3^4*3^4*3^4*3\equiv_{10} 1*1*1*3=3$.

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*3\equiv $

$1 + 36*6 + 3 \equiv $

$1 + 6*6 + 3\equiv $

$1 + 36 + 3\equiv $

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

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 $m\times n$ is the last digit of $b\times c$.

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

$1^{11} = 1$.

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

$3^{13} = 3^4\times 3^4 \times 3^4 \times 3$. $3^4 = 81$ so the last digit of $3^{13}$ is the last digit of $1\times1\times 1\times 3 = 3$>

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 $10$.

Answer 7

Note that $$11\equiv1\;\mathtt{mod}(10) \Rightarrow {11^{11}}\equiv {1^{11}}\equiv1 \;\mathtt{mod}(10)$$ Analugously $${12^{12}}\equiv{2^{12}}\equiv{(2^6)^2}\equiv{64^2}\equiv6\;\mathtt{mod}(10)$$ $${13^{13}}\equiv{(13^4)^2}*13\equiv{1^2}*3\equiv3\; \mathtt{mod}(10)$$ Thus $${11^{11}}+{12^{12}}+{13^{13}}\equiv1+6+3\equiv 0\; \mathtt{mod}(10)$$ $$\therefore10\mid \bigl({11^{11}}+{12^{12}}+{13^{13}}\bigr)$$

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

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

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

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

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

And we would be done.

PS:

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

Answer 8

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

Answer 9

Using the fact that $a^4\equiv1$ mod $5$ if $5\not\mid a$, we find

$$11^{11}+12^{12}+13^{13}\equiv \begin{cases}1+0+1\equiv0\mod 2\\ 1+1+3\equiv0\mod 5 \end{cases}$$

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

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

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!

Answer 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 $n^4\equiv 1$ 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.

Answer 11

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?

Answer 12

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

Answer 13

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 $10$

Answer 14

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 // $
$\displaystyle 40$