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- Feb 14, 2012

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- Thread starter anemone
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- Feb 14, 2012

- 3,909

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My solution:

\(\displaystyle \frac{10^{20000}}{10^{100}+3}=10^{19900}-3\cdot\frac{10^{19900}}{10^{100}+3}\)

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:

\(\displaystyle 10-7=3\)

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- Feb 14, 2012

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Thanks...andI found this to be a very interesting problem...

Well done,My solution:

\(\displaystyle \frac{10^{20000}}{10^{100}+3}=10^{19900}-3\cdot\frac{10^{19900}}{10^{100}+3}\)

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:

\(\displaystyle 10-7=3\)

I sure like your approach very very much!