smc

Define numbers $a,b,c,$ and $d$ to be the four numbers. In order to satisfy the following conditions, the system of equation should be constructed. (It doesn't matter which variable is which.) $$\frac{a+b+c}{3}+d=29$$ $$\frac{a+b+d}{3}+c=23$$ $$\frac{a+c+d}{3}+b=21$$ $$\frac{b+c+d}{3}+a=17$$ Adding all of the equations together, we get: $2(a+b+c+d)=90$. This means that $a+b+c+d=45$. We can determine that $\frac{a+b+c}{3}+d+16=a+b+c+d$. This, with some algebra, means that $\frac{1}{3}(a+b+c)=8$. $d$ must be $\boxed{21}$.