2004 AMC 10B Problem 13

Every coin thickness ends in a $5$ in the hundredths place, so consider the height’s last two decimal digits

For the height to be a whole number of mm, the number of coins must be a multiple of $4$

Measure every thickness in hundredths of a millimeter. The four possible thicknesses are $135,$ $155,$ $175,$ $195,$ all congruent to $15\pmod{20}.$ If the stack has $k$ coins and an integer height, its total in hundredths is divisible by $20,$ so $15k$ is divisible by $20.$ Therefore $k$ must be a multiple of $4.$

A stack of $4$ coins is at most $4(1.95) = 7.8$ mm, and a stack of $12$ coins is at least $12(1.35) = 16.2$ mm, so only $8$ coins can total $14$ mm.

Indeed, $8$ quarters give $8(1.75) = 14$ mm.

Thus, the correct answer is B.