2024 CSAT — Q5

Each straight planar cut along one axis multiplies the piece-count along that axis; cuts may be re-stacked between passes, so $c$ cuts along one direction produce $c+1$ slabs. To get $64$ identical small cubes we need a $a\times b\times c$ grid with $a\cdot b\cdot c = 64$ and we want to minimise the total cuts $(a-1)+(b-1)+(c-1)$.

For a fixed product, the sum of factors is minimised when the factors are as equal as possible. Since $64 = 4\times4\times4$,

Any less-balanced split costs more: $64 = 8\times8\times1 \Rightarrow 7+7+0 = 14$; $64 = 2\times4\times8 \Rightarrow 1+3+7 = 11$. The balanced $4\times4\times4$ is the optimum.

Answer: (b) 9.

Two ideas win this: (1) re-stacking — $n$ pieces along one axis cost $n-1$ cuts, never $n$; and (2) balance — for a fixed product, equal factors minimise the factor-sum, the AM–GM intuition. $64$ is the perfect cube $4^3$, so the symmetric split $4\times4\times4$ is forced as optimal: $3$ cuts on each of three axes $=9$. If $64$ were not a perfect cube you would pick the most-balanced factor triple. Recognising “minimise the sum of factors at fixed product” turns a fiddly 3-D visualisation into one line.