If $a$, $b$ and $c$ are positive integers such that $ab=432$, $bc=96$ and $c<9$, then the smallest possible value of $a+b+c$ is

We are given that $a$, $b$, and $c$ are positive integers such that:

$ab=432$

$bc=96$

with $c<9$.

Dividing the two equations gives:

$\frac{a}{c}=\frac{432}{96}=\frac{9}{2}$

Therefore, $a=4.5c$, which means $c$ must be an even positive integer because $a$ is an integer.

Since $c<9$, the possible values for the even positive integer $c$ are $2,4,6,8$.

Let us evaluate each case:

1. If $c=2$:
$a=9$
$b=\frac{96}{c}=48$
$a+b+c=9+48+2=59$

2. If $c=4$:
$a=18$
$b=\frac{96}{c}=24$
$a+b+c=18+24+4=46$

3. If $c=6$:
$a=27$
$b=\frac{96}{c}=16$
$a+b+c=27+16+6=49$

4. If $c=8$:
$a=36$
$b=\frac{96}{c}=12$
$a+b+c=36+12+8=56$

Comparing these sums, the smallest possible value is $46$.