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Let $x$ be a positive integer of the form $nnn\frac n2 \frac n2 \frac n2$ where $n$ is an even integer less than $30$.

Then $\frac x \pi$ is of the form $abab0c.\ ...$ where $a$ and $b$ are digits between $0$ and $9$, and $c$ is either $0$ or $1$. If $n>8$ carry digits as shown below.

For example, if $n=12$, then $x$ would be $(12)(12)(12)666=(12)(13)2666=(13)32666=1332666$, and $\frac x \pi=424200.76...$.

Proof by exhaustion is easy in this case as there is only a short list to test for. But why does this phenomenon occur? (it doesn't work for $3$, $3.14$ or even $3.141$ as divisors). Is there a shorter and more elegant proof to this?

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  • $\begingroup$ Yes, I realised that, too. My bad $\endgroup$ – Alvin Lepik Nov 26 '17 at 16:36
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Let $n$ be an even integer, $n<30,$ and let $$ x = 10^5 n + 10^4 n + 10^3 n + 10^2 \frac n2 + 10 \frac n2 + \frac n2. $$ Then $$ \frac x\pi = 10^5 a + 10^4 b + 10^3 a + 10^2 b + c + f $$ where $a,$ $b,$ and $c$ are integers, $0\leq a\leq 9,$ $0\leq b\leq 9,$ $0\leq c\leq 9,$ and $f$ is a real number with $0\leq f<1.$

Proof: Since $n$ is even, let $n=2m.$ Then $x=222111m,$ so $\frac x\pi \approx 70700.1272m = 7m \times 10100 + 0.1272m.$ If $1\leq m\leq 14,$ then $7\leq 7m\leq 98$ and $0\leq 0.1272m < 10.$ Let $7m=10a+b$ and $0.1272m = c+ f$ where $0\leq a\leq 9,$ $0\leq b\leq 9,$ and $0\leq c\leq 9.$ Then $0\leq f < 1$ and $\frac x\pi = 10^5 a + 10^4 b + 10^3 a + 10^2 b + c + f,$ as required. QED

Note that if $m\geq 15,$ then $7m>100,$ and $\frac x\pi$ has too many digits to match the pattern in the conjecture.

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