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- Feb 14, 2012
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What is the units digit of \(\displaystyle \left\lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor\)?
Thanks...andI found this to be a very interesting problem...![]()
Well done, MarkFL!My solution:
I began by writing:
\(\displaystyle \frac{10^{20000}}{10^{100}+3}=10^{19900}-3\cdot\frac{10^{19900}}{10^{100}+3}\)
Continuing in this manner, we will find:
\(\displaystyle \frac{10^{20000}}{10^{100}+3}= \sum_{k=1}^{199}\left((-3)^{k-1}10^{100(200-k)} \right)-(3)^{199}+\frac{9^{100}}{10^{100}+3}\)
Hence:
\(\displaystyle \left\lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor=\sum_{k=1}^{199}\left((-3)^{k-1}10^{100(200-k)} \right)-(3)^{199}\)
\(\displaystyle \left\lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor=m\cdot10^{100}-(3)^{199}\) where \(\displaystyle m\in\mathbb{N}\)
Now, we need only find the units digit of \(\displaystyle 3^{199}\) and subtract it from 10. Borrowing from my solution to last week's High School POTW...
Observing that:
\(\displaystyle 3^{4(1)-1}=27\)
\(\displaystyle 3^{4(2)-1}=2187\)
We may choose to state the induction hypothesis $P_n$:
\(\displaystyle 3^{4n-1}=10k_n+7\)
As the induction step, we may add:
\(\displaystyle 3^{4(n+1)-1}-3^{4n-1}=80\cdot3^{4n-1}=80\left(10k_n+7 \right)\)
to get:
\(\displaystyle 3^{4(n+1)-1}=80\left(10k_n+7 \right)+10k_n+7=10\left(8\left(10k_n+7 \right)+k_n \right)+7\)
If we make the recursive definition:
\(\displaystyle k_{n+1}\equiv8\left(10k_n+7 \right)+k_n\) where \(\displaystyle k_1=2\)
we then have:
\(\displaystyle 3^{4(n+1)-1}=10k_{n+1}+7\)
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:
\(\displaystyle 10-7=3\)