The locked chest

At home there is a small chest with three dials, each showing the digits zero to nine. That makes a thousand possible codes. Trying one every five seconds would take up to about an hour and twenty minutes.

A chest, $3$ dials, digits $0$ to $9$: $10^3=1000$ codes. At $5$ s per try that is up to $5000$ s, about $83$ min.

$(a,b,c)\in \{0,\dots ,9{\}}^3$: $1000\times 5\mathrm{s}=5000\mathrm{s}\approx 83\min$.

On the lid someone has scratched three clues: the digits grow from left to right; they add up to twelve; they multiply to forty-eight.

Three clues on the lid for the code $(a,b,c)$: first $a<b<c$, second $a+b+c=12$, third $a\cdot b\cdot c=48$.

$a<b<c,a+b+c=12,a\cdot b\cdot c=48$.

The first two clues alone leave ten codes, such as zero three nine and one five six. The third clue crosses out nine of them and keeps only two four six, since two plus four plus six is twelve and two times four times six is forty-eight.

Clues $1$ and $2$ alone leave $10$ codes, e.g. $(0,3,9)$ and $(1,5,6)$. Clue $3$ crosses out $9$ of them and keeps only $(2,4,6)$, since $2+4+6=12$ and $2\times 4\times 6=48$.

$S$ clues 1 and 2, $P$ product $48$: $|S|=10,|S\cap P|=1$, so $\exists !(a,b,c)=(2,4,6)$.

The chest opened at the first try. Inside was a smaller chest, and a note: same riddle, new numbers.

It opened at the first try. Inside: a smaller chest and a note, same riddle, new numbers.

Chest $c_1$ and $c_2\subset c_1$, with a note.