[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
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.
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
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.
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