Counting for ages

WLOG let $a > b > c > d$ .
We have $a = bcd$ and $a = b+c+d + 2$ .

Since $bcd = b + (c+1) + (d+1) < b + b + b = 3b$ , so $cd < 3$ which means $cd=2, c = 2, d = 1$ .
From there, solving $2b = b + 2 + 1 + 2$ gives us $b = 5$ .
Thus $a = bcd = 10$ .

Too long for a comment:

Set up the equations $a = bcd$ and $a = b + c + d + 2$ , again with $a > b > c > d$ without loss of generality.

Observe that for large values of $b,c,d$ , $bcd$ is much greater than $b+c+d+2$ . So if $a$ is odd, all three of $bcd$ must be odd. The first instance where this happens is when $b=5, c=3, d=1$ , but $5 \cdot 3 \cdot 1 > 5 + 3 + 1 + 2$ . So $a$ must be even. Similarly, $b,c,d$ cannot all be even as $6 \cdot 4 \cdot 2 > 6 + 4 + 2 + 2$ .

So in $b,c,d$ , there must be either 1 odd, 2 even, or 1 even, 2 odd. When there is 1 odd number and 2 even numbers, $a = b + c + d + 2$ implies that $a$ is odd., which is a contradiction. So there must be 1 even and 2 odd.

let a be the Oldest one so we get two equations

$a=b\times c \times d$ and $a = b+c+d+2$

smallest possible solution is $a = 10,b = 5,c=2$ and $d =1$

This is actually the same as Calvin Lin's solution but I'm trying to explain it more thoroughly.

Assume $d<c<b<a$ with each being a natural number. From the question we have $a=bcd$ and $a=b+c+d+2$ , and so $bcd=b+c+d+2$ .

Since $a,b,c,d$ are all natural numbers, from the assumption we get:

$c<b\implies c+1\leq b$ . Hence, $d<c<b \implies d+2\leq c+1\leq b$ .

Since $c<b$ and $d+2\leq b$ , we have $bcd=b+c+(d+2)<3b$ , so $cd<3$ that gives the only possible solution being $d=1,c=2$ .

Insert them to the equation above we get $2b=b+5$ so $b=5$ . Hence, $a=bcd=10$ .