Solution to Last 1000 digits puzzle

[This is a solution to the last 1000 digits puzzle earlier.]

A brief description of the problem:

What are the last 1000 digits of $$1 + 50 + 50^2 + \dots + 50^{999}$$ when written in base-10? A reasonably easy to compute (by a human) description is enough.

Solution

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The solution relies on the following fact:

If $a$ is relatively prime to $49$, then $a^{42}- 1$ is divisible by $49$: by applying Euler's theorem, as $\varphi(49) = 42$.

Now, the last $1000$ digits of $$1 + 50 + 50^2 + \dots + 50^{999}$$ are the same as the last $1000$ digits of $$S  = 1 + 50 + 50^2 + \dots + 50^{1007} = \frac{50^{1008} -1}{49}  =$$ $$\frac{(5^{1008}-1)10^{1008} + (10^{1008}-1)}{49}$$


Since $1008$ is divisible by $42$, $5^{1008}-1$ is divisible by $49$.

Thus

$$S = K*10^{1008} + \frac{10^{1008}-1}{49}$$

Thus the last $1000$ digits of $S$ are same as the last $1000$ digits of $$\frac{10^{1008}-1}{49}$$ which are the same as the last $1000$ digits of $$\left\lfloor\frac{10^{1008}}{49}\right\rfloor$$ .

Since $\dfrac{1}{49}$ is a repeating decimal with period $42$ , the last $1000$ digits can easily be computed:  $204081632653061224489795918367346938775510$ repeated.