MMXVI

2016-12-01, post № 149

mathematics, #all days of December

The idea is to only use the year’s digits — preferably in order — and mathematical symbols $(+,-,\cdot,\sqrt{},\lfloor\rfloor,\lceil\rceil,\dots)$ to create an equation that evaluates to a specific day of the month.
The 0th of December, 2016 would, for example, be $2\cdot 0\cdot 1\cdot 6$ , $2^0-1^6$ or $\lfloor\frac{2}{0+16}\rfloor$ .

$1=2\cdot 0+1^6$
$2=\sqrt{20-16}$
$3=(2+0)^{-1}\cdot 6$
$4=2^{0+\sqrt{\sqrt{16}}}$
$5=2\cdot 0-1+6$
$6=2\cdot 0\cdot 1+6$
$7=2\cdot 0+1+6$
$8=2+0+\sqrt[1]{6}$
$9=2+0+1+6$
$10=2+0!+1+6$
$11=-\lceil\sqrt{20}\rceil+16$
$12=2\cdot (0!-1+6)$
$13=-(2+0!)+16$
$14=-(2+0)+16$
$15=-(2\cdot 0)!+16$
$16=2\cdot 0+16$
$17=2^0+16$
$18=2+0+16$
$19=2+0!+16$
$20=\lceil\sqrt{20}\rceil\cdot\sqrt{16}$
$21=20+\lfloor\sqrt{\sqrt{\sqrt{16}}}\rfloor$
$22=(2+0!)!+16$
$23=(2+0!+1)!-\lfloor\sqrt{\sqrt{6}}\rfloor$
$24=20+\sqrt{16}$
$25=20-1+6$
$26=20+1\cdot 6$
$27=20+1+6$
$28=2+0-1+\lceil\sqrt{6!}\rceil$
$29=20+\lceil\sqrt{\lceil\sqrt{(1+6)!}\rceil}\rceil$
$30=(2+0!+1)!+6$
$31=\lceil\sqrt{\sqrt{\sqrt{\sqrt{20!}}}}\rceil +16$