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Make the 41 integers between -20 and 20 (-20, -19, ..., 0, 1, ...20) using only the four basic arithmetical operations, square root, the floor function, and, Surprise Surprise, exactly 3 π’s. No more and no less. No other digits or symbols are allowed. You can use as many square roots and floor functions as you like. Parenthesis are allowed. Unary minus sign is not allowed. Exponentiation is not allowed. Anything not explicitly allowed is disallowed.

The floor function for x, written $\lfloor x \rfloor$, is equal to the greatest integer smaller or equal to x. For example, $\lfloor \pi \rfloor = 3$, $\lfloor \pi \times \pi \rfloor = 9$ and $\lfloor -\pi \rfloor = −4$.

Shorter is better (i.e. try to find expressions with the minimal number of +, -, $\times$, $\div$, $\lfloor\rfloor$ and $\sqrt{}$'s.)

Note: 20 and -11 to -20 are very challenging. (0 to 10 are fun)

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4
  • $\begingroup$ Are we allowed to use things like $\pi^2$, or does that count as using a number ($2$) other than $\pi$? $\endgroup$ Dec 14, 2020 at 4:55
  • $\begingroup$ No! exactly three pi's and that it. $\endgroup$
    – pepster
    Dec 14, 2020 at 4:56
  • 2
    $\begingroup$ I was expecting a puzzle about actual pies... $\endgroup$
    – melfnt
    Dec 14, 2020 at 8:56
  • 1
    $\begingroup$ Isn't this open-ended? $\endgroup$ Dec 14, 2020 at 20:02

5 Answers 5

2
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Thanks to Retudin for $14,15,16,20$ and the inspiration necessary to find some others.

  • $-20=\left\lfloor\lfloor\pi\rfloor\div\left(\lfloor\sqrt{\pi}\rfloor-\sqrt{\sqrt{\sqrt{\pi}}}\right)\right\rfloor$

$-19$ not found yet

  • $-18=\left\lfloor\pi\div\left(\sqrt{\sqrt{\pi}}-\sqrt{\sqrt{\sqrt{\pi}}}\right)\right\rfloor$

  • $-17=\left\lfloor\sqrt{\sqrt{\pi}}\div\left(\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}} - \sqrt{\sqrt{\sqrt{\pi}}}\right)\right\rfloor$

$-16,-15,-14$ not found yet

  • $-13=\left\lfloor\pi\div\left(\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}} - \sqrt{\sqrt{\pi}}\right)\right\rfloor$

  • $-12=\left\lfloor\sqrt{\sqrt{\pi}}\div\left(\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}} - \sqrt{\sqrt{\sqrt{\pi}}}\right)\right\rfloor$

  • $-11=\left\lfloor\pi\div\left(\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}} - \sqrt{\sqrt{\pi}}\right)\right\rfloor$

  • $-10=\left\lfloor\pi\div\left(\lfloor\sqrt{\pi}\rfloor-\sqrt{\sqrt{\pi}}\right)\right\rfloor$

$-9$ not found yet

  • $-8=\lfloor\sqrt{\pi}\rfloor-\lfloor\pi\times\pi\rfloor$

  • $-7=\left\lfloor\sqrt{\pi}\div\left(\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}} - \sqrt{\sqrt{\pi}}\right)\right\rfloor$

  • $-6=\lfloor\pi\rfloor-\lfloor\pi\times\pi\rfloor$

  • $-5=\lfloor\sqrt{\pi}\rfloor-\lfloor\pi+\pi\rfloor$

  • $-4=\lfloor\pi-\pi-\pi\rfloor$

  • $-3=\pi-\pi-\lfloor\pi\rfloor$

  • $-2=(\pi\div\pi)-\lfloor\pi\rfloor$

  • $-1=\lfloor\sqrt{\pi}\rfloor-\lfloor\sqrt{\pi}\rfloor-\lfloor\sqrt{\pi}\rfloor$

  • $0=\lfloor\pi\div(\pi\times\pi)\rfloor$

  • $1=\lfloor\sqrt{\pi}\times\pi\div\pi\rfloor$

  • $2=\lfloor\pi\rfloor-(\pi\div\pi)$

  • $3=\lfloor\pi\times\pi\div\pi\rfloor$

  • $4=\lfloor\pi\rfloor+(\pi\div\pi)$

  • $5=\lfloor\sqrt{\pi\times\pi\times\pi}\rfloor$

  • $6=\lfloor\pi+\pi\rfloor\times\lfloor\sqrt{\pi}\rfloor$

  • $7=\lfloor\pi+\pi\rfloor+\lfloor\sqrt{\pi}\rfloor$

  • $8=\lfloor\sqrt{\sqrt{\pi}}\times(\pi+\pi)\rfloor$

  • $9=\lfloor\pi+\pi+\pi\rfloor$

  • $10=\lfloor\pi\times\pi\rfloor+\lfloor\sqrt{\pi}\rfloor$

  • $11=\lfloor\sqrt{\pi}\times(\pi+\pi)\rfloor$

  • $12=\lfloor\pi\times\pi\rfloor+\lfloor\pi\rfloor$

  • $13=\lfloor\sqrt{\sqrt{\pi}}\times\pi\times\pi\rfloor$

  • $14=\lfloor(\pi+\sqrt\pi)\times\lfloor\pi\rfloor\rfloor$

  • $15=\lfloor\lfloor\pi\rfloor\times\lfloor\pi\rfloor \times\sqrt\pi)\rfloor$

  • $16=\lfloor\pi\times\lfloor\pi\rfloor \times\sqrt\pi)\rfloor$

  • $17=\lfloor\pi\times\pi\times\sqrt{\pi}\rfloor$

  • $18=\lfloor\pi\rfloor\times\lfloor\pi+\pi\rfloor$

  • $19=\lfloor\pi\times(\pi+\pi)\rfloor$

  • $20=\left\lfloor\sqrt{\sqrt{\sqrt{\pi}}}\div\left(\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}} - \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}}}\right)\right\rfloor$

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6
  • $\begingroup$ Unary minus not allowed!! $\endgroup$
    – pepster
    Dec 14, 2020 at 5:12
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    $\begingroup$ @pepster :-/ you edited the question while I was answering ... OK, will try to fix this $\endgroup$ Dec 14, 2020 at 5:15
  • $\begingroup$ Also, there is (for example) a shorter expression for 0. $\endgroup$
    – pepster
    Dec 14, 2020 at 5:17
  • $\begingroup$ @pepster Updated my answer. Still haven't found all the numbers, but now all are correct with no unary minuses. $\endgroup$ Dec 14, 2020 at 5:33
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    $\begingroup$ "@pepster :-/ you edited the question while I was answering" Very annoying, indeed. $\endgroup$ Dec 14, 2020 at 5:41
1
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As OP moved the goal posts half-way through I can't be bothered to fix this ...


$0 = (\pi-\pi)\times\pi$
$\pm 1 = \pm (\pi^{\pi-\pi})$
$\pm 2 = \pm \frac{\pi+\pi}{\pi}$
$\pm 3 = \pm \lfloor \pi+\pi-\pi \rfloor$
$\pm 4 = \mp \lfloor \pi-\pi-\pi \rfloor$
$\pm 5 = \mp \lfloor -\frac {\pi}{\pi}-\pi \rfloor$
$\pm 6 = \pm \lfloor \pi \times \pi - \pi \rfloor$
$\pm 7 = \mp \lfloor \pi - \pi \times \pi \rfloor$
$\pm 8 = \pm \lfloor \frac {\lfloor \pi \rfloor^{\lfloor \pi \rfloor}} {\pi} \rfloor$
$\pm 9 = \pm \lfloor \frac {\pi^{\lfloor \pi \rfloor}} {\pi} \rfloor$
$\pm 10 = \mp \lfloor -\frac {\pi^{\lfloor \pi \rfloor}} {\pi} \rfloor$
$\pm 11 = \pm \lfloor \frac {\pi^{\pi}} {\pi} \rfloor$
$\pm 12 = \mp \lfloor -\frac {\pi^{\pi}} {\pi} \rfloor$
$\pm 13 = \pm \lfloor -\frac {\lfloor \pi \rfloor^{\lfloor \pi \rfloor}} {\lfloor -\sqrt{\pi} \rfloor} \rfloor$
$\pm 14 = \mp \lfloor \frac {\lfloor \pi \rfloor^{\lfloor \pi \rfloor}} {\lfloor -\sqrt{\pi} \rfloor} \rfloor$
$\pm 15 = \pm \lfloor \frac {\lfloor \pi \rfloor^{\pi}} {\lfloor -\sqrt{\pi} \rfloor} \rfloor$
$\pm 16 = \mp \lfloor \frac {\lfloor \pi \rfloor^{\pi}} {\lfloor -\sqrt{\pi} \rfloor} \rfloor$
$\pm 17 = \pm \lfloor \pi\pi \sqrt{\pi} \rfloor$
$\pm 18 = \mp \lfloor -\pi\pi \sqrt{\pi} \rfloor$
$\pm 19 = \pm \lfloor \pi(\pi+\pi) \rfloor$
$\pm 20 = \mp \lfloor -\pi(\pi+\pi) \rfloor$

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1
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'missing' positives
14-16

$14=\lfloor(\pi+\sqrt\pi)\times\lfloor\pi\rfloor\rfloor$

$15=\lfloor\lfloor\pi\rfloor\times\lfloor\pi\rfloor \times\sqrt\pi)\rfloor$

$16=\lfloor\pi\times\lfloor\pi\rfloor \times\sqrt\pi)\rfloor$

20

$20 = \lfloor\sqrt(\sqrt(\sqrt\pi)) / (\sqrt(\sqrt(\sqrt(\sqrt\pi))) - \sqrt(\sqrt(\sqrt(\sqrt(\sqrt(\sqrt\pi))))) \rfloor) $

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0
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The missing numbers (including some Rand also solved in the meantime)

$-9 = \lfloor \lfloor \sqrt{\pi} \rfloor - \pi \rfloor \times \lfloor \pi \rfloor$
$-10 = \lfloor \lfloor \lfloor \sqrt{\pi} \rfloor - \pi \rfloor \times \pi \rfloor$
$-12 = \lfloor\frac{\sqrt{\pi}}{\lfloor \sqrt{\pi} \rfloor - \sqrt{\sqrt{\sqrt{\pi}}}} \rfloor$
$-13 = \lfloor \frac{\sqrt{\pi}}{\lfloor \pi \rfloor - \pi}\rfloor$
$-14 = \lfloor\frac{\lfloor \sqrt{\pi} \rfloor}{\lfloor \sqrt{\pi} \rfloor - \sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}} \rfloor$
$-15 = \lfloor \frac{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}}{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}}}}}}} - \sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}}\rfloor$
$-16 = \lfloor \frac{ \sqrt{\pi} }{ \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}} - \sqrt{\sqrt{\sqrt{\pi}}}}\rfloor$
$-17 = \lfloor \frac{\lfloor \pi \rfloor}{ \sqrt{\sqrt{\sqrt{\pi}}} - \sqrt{\sqrt{\pi}}}\rfloor$
$-18 = \lfloor \frac{\pi}{ \sqrt{\sqrt{\sqrt{\pi}}} - \sqrt{\sqrt{\pi}}}\rfloor$
$-19 = \lfloor \frac{\sqrt{\sqrt{\pi}}}{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}}}}}}}- \sqrt{\sqrt{\sqrt{\sqrt{\pi}}}}} \rfloor$
$-20 = \lfloor\frac{\lfloor \pi \rfloor}{\lfloor \sqrt{\pi} \rfloor - \sqrt{\sqrt{\sqrt{\pi}}}} \rfloor$

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-1
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Those are "minimal" expressions. (some of you posted alternate forms with equal number of operations, but others were not minimal).

1

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1
  • 3
    $\begingroup$ Would it be possible to type these out instead of just showing a picture? $\endgroup$
    – bobble
    Dec 14, 2020 at 21:16

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