Any mathematics symbols and processes to use the numbers 3, 5, 6, and 7 once each to get 100.
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$\begingroup$ Once each? Or any combination? I have a couple using more than one of each ... :o) $\endgroup$– Will CrawfordFeb 14, 2018 at 21:10
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$\begingroup$ Clearly too broad. See also Handling underspecified “Make <this number> from <these digits>” questions on our Meta. $\endgroup$– Rubio ♦Feb 14, 2018 at 22:49
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9 Answers
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$\sqrt{7^6}-3^5$ ...need more characters
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$100=(57+3)/.6$ if decimal point is allowed.
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With the numbers in order:
$\frac{(3!)! - 5!}{6 \mod 7}$
EDIT: Added some "any processes" answers, some sillier than others:
$76 - 3 + 5 = 100$ (octal numerals)
$S(3*5*7-6)$ (successor function)
$$-\frac{\log\left[\left(\log\underbrace{\sqrt{\sqrt{\cdots\sqrt6}}}_{100}\right) / \log(3!)\right]}{\log{(7-5)}}$$ (generic method for obtaining any positive integer with 3,5,6 and 7, layout borrowed from ffao)
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$(3 + 7) ^ {\left\lceil \frac{6}{5} \right\rceil}$
answered Feb 14, 2018 at 20:10
Victor Stafusa - BozoNaCadeiaVictor Stafusa - BozoNaCadeia
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1$\begingroup$ no way, thought about getting 2 out of 6 and 5 forever, never thought of that $\endgroup$– GuestFeb 14, 2018 at 20:30
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If the "floor" function is acceptable:
$\left\lfloor 3\times5\times6.7 \right\rfloor$
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$7!!-5×6/3!$ using double factorial
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35+67-2 $%The question doesn't say we can't use other digits$
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$\begingroup$ @Catija It's a "loophole" answer: the question doesn't say only those numbers. $\endgroup$ Feb 14, 2018 at 22:07
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${{6}\choose{3}} * 5 + \dot{7}$
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$\begingroup$ Could you explain a bit more what each step here means? I get that C(6,3) is combinations, but what's the apostrophe supposed to be? Is it a derivative? $\endgroup$– DqwertyCFeb 14, 2018 at 22:03
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1$\begingroup$ Yeah, that's it... Derivation of constant trick $\endgroup$ Feb 14, 2018 at 22:11
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$\begingroup$ I've updated it to use MathJax, if that's alright with you. $\endgroup$– DqwertyCFeb 14, 2018 at 22:23
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$$ \left \lceil 357 * \sin(-6) \right \rceil = 100$$
