Why are $5$ and $6$ (and numbers ending with these respective last digits) the only (nonzero, non-one) numbers such that no matter what (integer) power you raise them to, the last digit stays the same? (by the way please avoid modular arithmetic) Thanks!
[Math] What makes $5$ and $6$ so special that taking powers doesn’t change the last digit
algebra-precalculuselementary-number-theory
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Best Answer
The problem is solving $x^2\equiv x\pmod{10}$, or $x(x-1)\equiv 0\pmod{10}$, which means finding integers $x$ such that $10$ is a factor of $x(x-1)$. For that to hold, either $x$ or $x-1$ must be a multiple of $5$, which means the last digit of $x$ is $0,1,5,$ or $6$. Then it is a simple verification that the equation holds in each of these cases.
Rephrased without "$\pmod{10}$" notation, this could be expressed as follows. We are looking for integers $x$ such that $x$ and $x^2$ have the same last digit. This is the same as saying that the last digit of $x^2-x=x(x-1)$ is $0$. That means that $x(x-1)$ is a multiple of $10$. See above for the rest.