# Find the Units Digit of An Expression

#### anemone

##### MHB POTW Director
Staff member
What is the units digit of $$\displaystyle \left\lfloor \frac{10^{20000}}{10^{100}+3} \right\rfloor$$?

#### MarkFL

Staff member
I found this to be a very interesting problem...

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:

$$\displaystyle 10-7=3$$

#### anemone

##### MHB POTW Director
Staff member
I found this to be a very interesting problem...
Thanks...and

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:

$$\displaystyle 10-7=3$$
Well done, MarkFL!

I sure like your approach very very much!