Decomposing a large block $7\times 4 \times 5$ into small ones $1\times 1 \times 1,\ 2\times 2 \times 2,\ 3\times 3 \times 3$ Consider a block of size $7\times 4 \times 5$. We want to make this with cubes of size $1\times 1 \times 1,\ 2\times 2 \times 2,\ 3\times 3 \times 3$. If $x,\ y,\ z$ cubes are needed respectively, then what is the smallest number $x$ ?
Proof : Here $z$ must be in $\{0,1,2\}$. $(x,y,z) = (38,6,2),\ (33,10,1),\ (44,12,0)$ are eamples. Hence I think that $x=33$ is the smallest. How can we prove this ?

As noted in the comments, there is a valid configuration with $(x,y,z)=(22,8,2)$.

Claim:$\;$Every valid configuration $(x,y,z)$ has $x\ge 22$.

Proof:

Suppose $(x,y,z)$ is a valid configuration.

Summing the volumes, we get $x+8y+27z=140$, hence $x\equiv 4-3z\;(\text{mod}\;8)$.

As you noted, we must have $z\in\{0,1,2\}$.

Place the large block so it has horizontal base $7{\times}4$ and height $5$, and regard it as comprised of $28$ vertical $1{\times}1{\times}5$ columns.

Then each of the $28$ vertical columns either contains a $1$-cube, or passes through the interior of exactly one $3$-cube.

It follows that $x\ge 28-9z$.

Consider $3$ cases . . .

Case $(1)$:$\;z=0$.

Then from $x\ge 28-9z$, we get $x\ge 28$.

Case $(2)$:$\;z=1$.

Then from $x\ge 28-9z$, we get $x\ge 19$, and from $x\equiv 4-3z\;(\text{mod}\;8)$,we get $x\equiv 1\;(\text{mod}\;8)$, hence $x\ge 25$.

Case $(3)$:$\;z=2$.

Now regard the large block as comprised of $35$ horizontal $1{\times}1{\times}4$ rows.

Since $z=2$, exactly $18$ of the $35$ horizontal rows pass through the interior of a $3$-cube.

Necessarily each of those $18$ rows also contains a $1$-cube, hence $x\ge 18$.

But then from $x\equiv 4-3z\;(\text{mod}\;8)$,we get $x\equiv 6\;(\text{mod}\;8)$, so $x\ge 22$.

Thus in all cases, we have $x\ge 22$.
