Question

Suppose $a, b, c$ are three distinct natural numbers, such that $3ac = 8(a + b)$. Then, the smallest possible value of $3a + 2b + c$ is:

Hint:

When you have a Diophantine equation (integer solutions) and need to minimize an expression, first express one variable in terms of the others, then systematically test small integer values under the constraints (like distinctness and positivity).

Answer 1

Correct Answer: 12

Step 1: Rearranging, \(a = \dfrac{8b}{3c-8}\). Since \(b\) carries the smallest weight in \(3a+2b+c\), try small values of \(b\) first.
Step 2: For \(b=1\): \(a = \dfrac{8}{3c-8}\), so \(3c-8\) must divide \(8\). Taking \(3c-8=4\) gives \(c=4,\,a=2\) (all distinct), so \(3a+2b+c = 6+2+4=12\).
Step 3: For \(b=2\): \(a=\dfrac{16}{3c-8}\); the best valid case is \(3c-8=16\), giving \(c=8,\,a=1\), so the sum is \(3+4+8=15\), which is larger.
Step 4: Larger \(b\) values only push the sum higher, so \(b=1\) with \((a,c)=(2,4)\) is optimal.
\[ \boxed{3a+2b+c=12} \]