Let there be \(N\) ball bearings of radius \(r\) and density \(\rho\).   This means that the total volume of the ball bearings is \((4/3)\pi r^3 N\) and the volume of the same weight of water is \(\rho\) times this.  The drum radius is \(Nr\), giving the volume of water in a drum height of \(h\) as \(\pi r^2 N^2 h\).

When the ball bearings are dropped into the water the change in its height \(h_1\) is given by:\[\pi r^2 N^2 h_1 = \frac{4}{3}\pi r^3N\] which gives \(h_1=4r/3N\).

When the ball bearings are placed in the boat, the boat will sink until the increased pressure is sufficient to support their weight. If it sinks by a length \(h\) we can equate the increased water pressure to the pressure on the bottom of the boat created by the ball bearings:\[h^{‘} g=\frac{(4/3)\pi r^3N\rho g}{\pi r^2 N^2/10}\]where \(g\) is the acceleration due to gravity. This gives \(h^{‘}=(40r\rho/3N)\). When the bearings are taken out the boat rises and the displaced water spreads across the whole drum and the height drops to \(h^{”}\) such that \(h^{”} A=(9/10)h^{‘}A\) where \(A\) is the area of the drum’s base. So the height change when the ball bearings are loaded is:\[h^{‘} – h^{”} = \frac{4r\rho}{3N}\]Equating the difference in these two heights to the ball radius now gives: \[3N=4(\rho-1)\]Since \(\rho\) is a single digit number this show immediately that \(\rho\) is either 4 or 7 and \(n\) is either 4 or 8.

It appears that the teaser author wrongly assumed that the displaced water would end up in the annular ring of water round the boat so that:\[(9/10)\pi r^2Nh=(4/3)\pi r^3 N \rho\]This gives the change in the height of the water as \((40r\rho/27N)\) which leads to a different equation:\[27N=4(10\rho-9)\]with the single solution of \(\rho=9\) leading to \(n=12\) as the intended solution. I also made the same mistake in my initial solution.