# Advent Puzzles 2014: puzzle 23 solution

Most numbers have several solutions. One solution for each possible total is given below. $1=\frac{1}{2}\times\frac{1}{3}\times2\times3$ $2=(\frac{1}{2}\times2)+(\frac{1}{3}\times3)$ $3=(2\div\frac{1}{2})-(3\times\frac{1}{3})$ $4=(2\div\frac{1}{2})\times3\times\frac{1}{3}$ $5=(3\div\frac{1}{3})-(2\div\frac{1}{2})$ $6=(3-2)\div\frac{1}{2}\div\frac{1}{3}$ $7\mbox{ cannot be made.}$ $8=(3\div\frac{1}{3})-(2\times\frac{1}{2})$ $9=(3\div\frac{1}{3})\times2\times\frac{1}{2}$ $10=(3\div\frac{1}{3})+(2\times\frac{1}{2})$ $11=((2\div\frac{1}{2})-\frac{1}{3})\times3$ $12=(3\div\frac{1}{2})+(2\div\frac{1}{3})$ $13=(2\div\frac{1}{2})+(3\div\frac{1}{3})$ $14=((3\div\frac{1}{3})-2)\div\frac{1}{2}$ $15=((3-\frac{1}{2})\times2)\div\frac{1}{3}$ $16=(3\div(\frac{1}{2}\times\frac{1}{3}))-2$ $17=((3\div\frac{1}{3})-\frac{1}{2})\times2$ $18=((2\div\frac{1}{3})+3)\div\frac{1}{2}$ $19=((3\div\frac{1}{3})+\frac{1}{2})\times2$ $20=(3\div(\frac{1}{2}\times\frac{1}{3}))+2$