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Is it possible to obtain the digits from 0 to 9 starting from 2019 and using its digits in the same order, together with the usual operations +, *, -, /, concatenation of digits, and the less usual operators ^, !, sqrt(), int()? For example, 1 = 20-19. Unary minus is allowed too.

I manage to use only basic operations and elevation to a power for all digits except 4 and 5, but maybe somebody will do better!

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1 Answer 1

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$0 = 2 \cdot 0 \cdot 1 \cdot 9$

$1 = 20 - 19$

$2 = 2^0 + 1^9$

$3 = 2+0+1^9$

$4 = \lfloor \sqrt{20} \rfloor + \lfloor 1/9 \rfloor$

$5 = 2 + 0 \cdot 1 + \sqrt{9}$

$6 = -(2+0+1) + 9$

$7 = -(2+0 \cdot 1) + 9$

$8 = -(2 \cdot 0 + 1) + 9$

$9 = -(2 \cdot 0 \cdot 1) + 9$

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    $\begingroup$ @deepthought Fixed. $\endgroup$ Commented Dec 23, 2018 at 21:29
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    $\begingroup$ Nice! 4 may also be $2\cdot 0 + 1 + \sqrt 9$ $\endgroup$
    – mau
    Commented Dec 23, 2018 at 21:48
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    $\begingroup$ A few more for 4 and 5: $\lfloor\sqrt{\sqrt{20\times19}}\rfloor=4$, $\lceil\sqrt{\sqrt{20\times19}}\rceil=5$ , $\lceil\sqrt{2+0+1+9}\rceil=4$ , $\lfloor\sqrt{2+0+19}\rfloor=4$ , $\lceil\sqrt{2+0+19}\rceil=5$ $\endgroup$
    – JMP
    Commented Dec 23, 2018 at 22:02
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    $\begingroup$ and $\lfloor\sqrt{201}\rfloor-9=5$ $\endgroup$
    – JMP
    Commented Dec 23, 2018 at 22:05
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    $\begingroup$ Fun fact: $-\lfloor -x \rfloor = \lceil x \rceil.$ $\endgroup$ Commented Dec 23, 2018 at 22:55

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