MHB Find the Units Digit of An Expression

[anemone]

What is the units digit of $$\left\lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor$$?

 

[MarkFL]

I found this to be a very interesting problem...(Clapping)

My solution:

Spoiler

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:

$$10-7=3$$

 

[anemone]

I found this to be a very interesting problem...(Clapping)

Thanks...and

My solution:

Spoiler

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:

$$10-7=3$$

Well done, MarkFL!(Clapping)

I sure like your approach very very much!:)