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

Question

Below are the steps I have done so far.

Find $a$ in terms of $c$

$$ ab = 432 \\ bc = 96 \\ \frac{ab}{bc} = \frac ac = \frac 92 \implies a = 4.5c. $$

Find $b$ in terms of $c$

$$ ab = 432 \\ \implies 4.5c \cdot b = 432 \\ \implies b = \frac{96}c. $$

Substituting, we get

$$ a + b + c = 4.5c + \frac{96}c + c = 5.5c + \frac{96}c. $$

How should I proceed from here?

Answer 1

The constraint on $c$ mentioned in the other answers can be sharpened by noting that $a =4.5c$ is only an integer when $c$ is even, so you only have to consider $c \in \{2,4,6,8\}$. Unfortunately $b= \frac{96}c$ does not permit further constraint, so you have to test those four values.

If you're permitted calculus, you can observe that $f(c) = 5.5c + \frac{96}c$ has a single turning point and reaches a global minimum for real $c$ at $c = \sqrt{\frac{192}{11}} \approx 4$, so this is very likely to be the correct solution based on the "nice behaviour" of such functions. But you really do have to test at least $c=4$ and $c=6$ (either side of the turning point) to be certain.

Answer 2

The factors of $96$ that are less than $9$ are $\{1,2,3,4,6,8\}$

Plug these values of $c$ into your equation above to find the set of $a+b+c$

Answer 3

$bc=96$ and $c<9$ constrains $c$ to be one of $1,2,3,4,6,8$. Since $\gcd(432,96)=48$, $c$ cannot be $1$ or $3$, leaving four cases: $(a,b,c)=(9,48,2),(18,24,4),(27,16,6),(36,12,8)$. Of these, $(18,24,4)$ achieves the smallest $a+b+c$ of $46$.