smc

Let there be $n$ integers on the list. The list of $n$ integers has mean $22$, so the sum of the integers is $22n$. Replacing $m$ with $m+10$ will increase the sum of the list from $22n$ to $22n+10$. The new mean of the list is $24$, so the new sum of the list is also $24n$. Thus, we get $22n+10=24n$, leading to $n=5$ numbers on the list. If there are $5$ numbers on the list with mode $32$ and smallest number $10$, then the list is $\{10,x,m,32,32\}$ Since replacing $m$ with $m-8$ gives a new median of $m-4$, and $m-4$ must be on the list of $5$ integers since $5$ is odd, $x=m-4$, and the list is now $\{10,m-4,m,32,32\}$ The sum of the numbers on this list is $22n=22\cdot 5=110$, so we get: $10+m-4+m+32+32=110$ $70+2m=110$ $m=20$, giving answer $\boxed{20}$. The original list is $\{10,16,20,32,32\}$, with mean $\frac{10+16+20+32+32}{5}=22$ and median $20$ and mode $32$. The second list is $\{10,16,30,32,32\}$, with mean $\frac{10+16+30+32+32}{5}=24$ and median $m+10=30$. The third list is $\{10,16,12,32,32\}\to \{10,12,16,32,32\}$ with median $m-4=16$.