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Without being proficient in math at all, I have figured out, by looking at series of numbers, that $6$ in the $n$-th power always seems to end with the digit $6$.
Anyone here willing to link me to a proof?
I've been searching google, without luck, probably because I used the wrong keywords.
We can prove it using mathematical induction.
Claim: $6^n\equiv 6\bmod 10$ for all $n\in\mathbb{N}$ (the symbol $\mathbb{N}$ denotes the natural numbers, and $\bmod 10$ means we are using modular arithmetic with a modulus of 10).
Base case (i.e., showing it's true for $n=1$): $$6^1\equiv 6\bmod 10\qquad\checkmark$$
Induction step (i.e., showing that, if it is true for $n=k$, then it is true for $n=k+1$):
$$6^k\equiv 6\bmod 10\implies 6^{k+1}\equiv 6^k\cdot 6\equiv6\cdot 6\equiv 36\equiv 6\bmod 10\qquad\qquad\checkmark$$
If you multiply any two integers whose last digit is 6, you get an integer whose last digit is 6: $$ \begin{array} {} & {} & {} & \bullet & \bullet & \bullet & \bullet & \bullet & 6 \\ \times & {} & {} &\bullet & \bullet & \bullet & \bullet & \bullet & 6 \\ \hline {} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & 6 \end{array} $$ (Get 36, and carry the "3", etc.)
To put it another way, if the last digit is 6, then the number is $(10\times\text{something}) + 6$. So $$ \begin{align} & \Big((10\times\text{something}) + 6\Big) \times \Big((10\times\text{something}) + 6\Big) \\ = {} & \Big((10\times\text{something})\times (10\times\text{something})\Big) \\ & {} + \Big((10\times\text{something})\times 6\Big) + \Big((10\times\text{something})\times 6\Big) + 36 \\ = {} & \Big(10\times \text{something}\Big) +36 \\ = {} & \Big(10\times \text{something} \Big) + 6. \end{align} $$
HINT $\rm\ \ 6-1\ |\ 6^k-1,\ $ so $\rm\:\ 2,5\ |\ 6^n-6\ \Rightarrow\ 10\ |\ 6^n - 6\:,\ $ i.e. $\rm\ 6^n\ =\ 6 + 10\ k\:$ for $\rm\:k\in\mathbb Z\:.$
Alternatively: $\rm\ mod\ 10:\ \ 6^n\equiv 6\ $ since it is $\rm\ 0^n \equiv 0\pmod 2,\ \ 1^n \equiv 1\pmod 5$
Similarly odd $\rm\:b\: \Rightarrow\: (b+1)^n\equiv b+1\pmod{2\:b}\:,\:$ so $\rm\:(b+1)^n\:$ has last digit $\rm\:b+1\:$ in radix $\rm\:2\:b\:.$
NOTE how modular arithmetic reduces the induction to the trivial inductions $\rm\ 0^n = 0,\ 1^n = 1\:.$ This is a prototypical example of the sort of simplification afforded by reducing arithmetical problems to their counterparts in the simpler arithmetical rings of integers $\rm\:(mod\ m)\:.\:$
This follows from the more general result that the product of two numbers ending with digit 6 also ends with digit 6. This can be proved in an elementary way: $$(10x+6)\cdot(10y+6) = 100xy + 60x +60y + 36 = 10(10xy+6x +6y +3) + 6 = 10z+6$$
Of course, avoiding all these letters is what congruences are all about.
$6 \times 6 \equiv 6 \pmod{10}$.
Or, more elementarily put, think back to the pen-and-paper multiplication algorithm. When you multiply something by 6, the only part of the original number that can affect the last digit of the result is the last digit of the original. If you start with something that ends in 6, you get 36 for the last position, write 6 down and carry the 3. But no matter what happens after the carry, it cannot affect the final 6 that you've already produced.
I also think that this is true, because $6\times 6=36$, meaning that any ending $6$ digit results in an ending $6$ digit through multiplication.
