Finding unit digit of $f(10)$

Question

If we define $$f(x)=\left\lfloor \frac {x^{2x^4}}{x^{x^2}+3}\right\rfloor$$ and we have to find unit digit of $f(10)$

I had tried approximation, factorization and substitutions like $x^2=u$ but it proved of no use. Moreover the sequential powers are feeling the hell out of me. Can someone please provide me with some hints

Answer 1

First substitute in $10$ for $x$ $$f(10)=\left\lfloor \frac {10^{2\cdot 10^4}}{10^{10^2}+3}\right\rfloor\\ =\left\lfloor \frac {10^{20000}}{10^{100}+3}\right\rfloor$$ Now ask Alpha
or ask Python

and the answer is $3$

Added in response to the comment: You can do long division in base $10^{100}$. Unfortunately the numerator still has $200$ digits, so it will be a long haul. The denominator is a simple $13$. I suspect you are intended to write $$f(10)=\left\lfloor \frac {10^{2\cdot 10^4}}{10^{10^2}+3}\right\rfloor\\ =\left\lfloor 10^{19900}\frac {10^{100}}{10^{100}+3}\right\rfloor\\ =\left\lfloor 10^{19900}\frac {1}{1+3\cdot 10^{-100}}\right\rfloor\\ =\left\lfloor 10^{19900}(1-3\cdot 10^{-100}+(3\cdot 10^{-100})^2-(3\cdot 10^{-100})^3+\ldots )\right\rfloor$$ and note that all the terms with exponents less than $199$ get too many zeros from the $10^{19900}$ to matter, then evaluate the term with exponent $199$. Then note that the term with exponent $200$ doesn't carry and is positive, so you only care about the term with exponent $199$. We have $-(3^{199})\equiv -7 \equiv 3 \pmod {10}$ so the answer is $3$.

Answer 2

When $x=10$, $\displaystyle \frac {x^{2x^4}}{x^{x^2}+3}=\frac {10^{20000}}{10^{100}+3}$.

Let $y=10^{100}$. $\displaystyle \frac {10^{20000}}{10^{100}+3}=\frac{y^{200}}{y+3}$.

When $y^{200}$ (as a polynomial) is divided by $y+3$, the remainder is $(-3)^{200}=3^{200}$.

So, $\displaystyle \frac{y^{200}}{y+3}=Q(y)+\frac{3^{200}}{y+3}$ for some polynomial $Q(y)$.

Put $y=0$, $Q(0)=-3^{199}=(-1)(81)^{49}(27)\equiv 3$ (mod $10$).

Note that $\displaystyle \frac{3^{200}}{10^{100}+3}=\frac{9^{100}}{10^{100}+3}<1$.

$$ \frac {10^{20000}}{10^{100}+3}=Q(10^{100})=10^{100}\times \textrm{some positive integer}+3+\frac{9^{100}}{10^{100}+3}$$

The unit digit of $\displaystyle \left\lfloor\frac {10^{20000}}{10^{100}+3}\right\rfloor$ is $3$.