The discussion centers on finding the units digit of the expression $$\left\lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor$$. Participants express enthusiasm for the problem, with one user, MarkFL, receiving commendation for their approach. The problem involves understanding the behavior of large powers of ten and their division by polynomial expressions. The conclusion emphasizes the importance of mathematical reasoning in determining the units digit of complex expressions.
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MarkFL said:I found this to be a very interesting problem...(Clapping)
MarkFL said:My solution:
I began by writing:
$$\frac{10^{20000}}{10^{100}+3}=10^{19900}-3\cdot\frac{10^{19900}}{10^{100}+3}$$
Continuing in this manner, we will find:
$$\frac{10^{20000}}{10^{100}+3}= \sum_{k=1}^{199}\left((-3)^{k-1}10^{100(200-k)} \right)-(3)^{199}+\frac{9^{100}}{10^{100}+3}$$
Hence:
$$\left\lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor=\sum_{k=1}^{199}\left((-3)^{k-1}10^{100(200-k)} \right)-(3)^{199}$$
$$\left\lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor=m\cdot10^{100}-(3)^{199}$$ where $$m\in\mathbb{N}$$
Now, we need only find the units digit of $$3^{199}$$ and subtract it from 10. Borrowing from my solution to last week's High School POTW...
Observing that:
$$3^{4(1)-1}=27$$
$$3^{4(2)-1}=2187$$
We may choose to state the induction hypothesis $P_n$:
$$3^{4n-1}=10k_n+7$$
As the induction step, we may add:
$$3^{4(n+1)-1}-3^{4n-1}=80\cdot3^{4n-1}=80\left(10k_n+7 \right)$$
to get:
$$3^{4(n+1)-1}=80\left(10k_n+7 \right)+10k_n+7=10\left(8\left(10k_n+7 \right)+k_n \right)+7$$
If we make the recursive definition:
$$k_{n+1}\equiv8\left(10k_n+7 \right)+k_n$$ where $$k_1=2$$
we then have:
$$3^{4(n+1)-1}=10k_{n+1}+7$$
We have derived $P_{n+1}$ from $P_n$ thereby completing the proof by induction.
Thus, we find the units digit of the original expression is:
$$10-7=3$$