Advent Puzzles 2014: puzzle 23 solution

Advent puzzle 23: Use all of 2, 3, ½ and ⅓, with any of + − × ÷ ( ), to make all totals from 1 to 20. One total can't be made. Which one?

— Portsmouth MathsJam (@MathsJamPOR) December 23, 2014

Most numbers have several solutions. One solution for each possible total is given below.

$1=\frac{1}{2}\times\frac{1}{3}\times2\times3$

$2=(\frac{1}{2}\times2)+(\frac{1}{3}\times3)$

$3=(2\div\frac{1}{2})-(3\times\frac{1}{3})$

$4=(2\div\frac{1}{2})\times3\times\frac{1}{3}$

$5=(3\div\frac{1}{3})-(2\div\frac{1}{2})$

$6=(3-2)\div\frac{1}{2}\div\frac{1}{3}$

$7\mbox{ cannot be made.}$

$8=(3\div\frac{1}{3})-(2\times\frac{1}{2})$

$9=(3\div\frac{1}{3})\times2\times\frac{1}{2}$

$10=(3\div\frac{1}{3})+(2\times\frac{1}{2})$

$11=((2\div\frac{1}{2})-\frac{1}{3})\times3$

$12=(3\div\frac{1}{2})+(2\div\frac{1}{3})$

$13=(2\div\frac{1}{2})+(3\div\frac{1}{3})$

$14=((3\div\frac{1}{3})-2)\div\frac{1}{2}$

$15=((3-\frac{1}{2})\times2)\div\frac{1}{3}$

$16=(3\div(\frac{1}{2}\times\frac{1}{3}))-2$

$17=((3\div\frac{1}{3})-\frac{1}{2})\times2$

$18=((2\div\frac{1}{3})+3)\div\frac{1}{2}$

$19=((3\div\frac{1}{3})+\frac{1}{2})\times2$

$20=(3\div(\frac{1}{2}\times\frac{1}{3}))+2$

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