imc

All numbers in this solution will be in hundredths of a millimeter. The thinnest coin is the dime, with thickness $135$. A stack of $n$ dimes has height $135n$. The other three coin types have thicknesses $135+20$, $135+40$, and $135+60$. By replacing some of the dimes in our stack by other, thicker coins, we can clearly create exactly all heights in the set $\{135n,135n+20,135n+40,\dots ,195n\}$. If we take an odd $n$, then all the possible heights will be odd, and thus none of them will be $1400$. Hence $n$ is even. If $n<8$ the stack will be too low and if $n>10$ it will be too high. Thus we are left with cases $n=8$ and $n=10$. If $n=10$ the possible stack heights are $1350,1370,1390,\dots$, with the remaining ones exceeding $1400$. Therefore there are $\boxed{8}$ coins in the stack. Using the above observation we can easily construct such a stack. A stack of $8$ dimes would have height $8\cdot 135=1080$, thus we need to add $320$. This can be done for example by replacing five dimes by nickels (for $+60\cdot 5=+300$), and one dime by a penny (for $+20$).