-6
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Use the numbers 2, 0, 1, and 8 (ALL of them...only ONE time each) to make every integer from 33 to 100.

  • Allowed operations: +, -, x, ÷, ! (factorial), exponentiation, square root and Parentheses
  • No specific order is needed
  • The modulus operator is not allowed
  • Rounding is not allowed (e.g. 201/8=25)
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  • 1
    $\begingroup$ Welcome to Puzzling.SE! We've been seeing quite a few of these posts lately, with some of them defining clear constraints or asking specific numbers (which are generally at least not ill-received) and some, like this one, with little to no constraints or asking for help simply solving the entire problem rather than specific ones (like this question), which are generally not well-received). See also my answer abusing the "no constraints". If you'd add some, or ask for some specific numbers, you might get a better reception. $\endgroup$ – Lolgast Jan 9 '18 at 10:01
  • $\begingroup$ I think I already read the exact same text here in another "puzzle". $\endgroup$ – Tweakimp Jan 9 '18 at 10:22
3
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You can generate any number X not just 1 to 100, like this
$x = \log_\sqrt{{\frac{2}{8}}}\left({\lg\underbrace{\sqrt{\sqrt{\dots\sqrt{\frac{0!}{.1}\,}\,}\,}}_\text{x square roots}}\right)$

Explanation

It works because...
$\log_\sqrt{{\frac{2}{8}}}\left({\lg\underbrace{\sqrt{\sqrt{\dots\sqrt{\frac{0!}{.1}\,}\,}\,}}_\text{x square roots}}\right)$ = $\log_\sqrt{{\frac{1}{4}}}\left({\lg\underbrace{\sqrt{\sqrt{\dots\sqrt{\frac{1}{\frac{1}{10}}\,}\,}\,}}_\text{x square roots}}\right)$ =
$\log_{\frac{1}{2}}\left({\lg\underbrace{\sqrt{\sqrt{\dots\sqrt{10\,}\,}\,}}_\text{x square roots}}\right)$ = $\log_{\frac{1}{2}}\left({\lg{10^{\frac{1}{2^x}}}}\right)$ = $\log_{\frac{1}{2}}\frac{1}{2^x}$ = $x$

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  • $\begingroup$ Indeed another combination of function abusing the absence of constraints. As a note, since here have been comments on this on other questions... the "x square roots" does not introduce another number, it indicates an amount. There are 3 square roots in $\sqrt{\sqrt{\sqrt{64}}}$ but that doesn't mean a 3 is used. $\endgroup$ – Lolgast Jan 9 '18 at 10:08
1
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Well, if we can use any mathematical operation...

I'll just use the successor function:
1 = $1+0*2*8$
2 = $S(1)+0*2*8$
3 = $S(S(1))+0*2*8$
4 = $S(S(S(1)))+0*2*8$
... I guess you get the point, this question is not constrained enough.

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  • $\begingroup$ Oh come on, you could at least have used $S(2) + 0 * 1 * 8$ :-) $\endgroup$ – Bass Jan 9 '18 at 10:34
  • $\begingroup$ @bass that doesn't make 1 :P I figured I'd just stick with the ordering afterwards. $\endgroup$ – Lolgast Jan 9 '18 at 10:38

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