Advent Puzzles 2014: puzzle 23 solution

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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