CAT 2022 Slot 1 · Quant · Question 4

Each equation has a common factor waiting to be taken out.

The first gives y\left( x+z \right)=19.

Since 19 is prime, the two factors must be 1 and 19 in some order.

Now x and z are natural numbers, so x+z\ge 2 — it cannot be 1.

Therefore y=1 and x+z=19.

The second equation gives z\left( y+x \right)=51, and with y=1 this is z\left( 1+x \right)=51.

Substituting z=19-x, \left( 19-x \right)\left( 1+x \right)=51.

Expanding, 19+18x-{x}^{2}=51, which rearranges to {x}^{2}-18x+32=0.

Factorising, \left( x-2 \right)\left( x-16 \right)=0, so x=2 or x=16.

When x=2, z=17 and xyz=2\times 1\times 17=34.

When x=16, z=3 and xyz=16\times 1\times 3=48.

The minimum is 34.

Shortcut. The prime 19 does all the work: it forces y=1 immediately, after which the problem is one quadratic. The two roots are symmetric about 9, so the more lopsided split gives the smaller product.