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

We are given \[ 3ac = 8(a + b). \]
Step 1: Express $b$ in terms of $a$ and $c$. \[ 3ac = 8a + 8b \;\Rightarrow\; 8b = 3ac - 8a = a(3c - 8) \;\Rightarrow\; b = \frac{a(3c - 8)}{8}. \] Since $a,b,c$ are natural numbers, $b$ must be a positive integer; hence $a(3c-8)$ must be divisible by $8$ and $3c-8>0$, so $c\ge 3$. 
Step 2: Expression to minimize. Let \[ S = 3a + 2b + c. \] Substitute the expression for $b$: \[ S = 3a + 2\cdot\frac{a(3c-8)}{8} + c = 3a + \frac{a(3c-8)}{4} + c = \frac{12a + 3ac - 8a}{4} + c = \frac{4a + 3ac}{4} + c = a\Bigl(1 + \frac{3c}{4}\Bigr) + c. \] We now test small integer values of $c$ that make $b$ integral and keep $a,b,c$ distinct. Case 1: $c = 3$ \[ b = \frac{a(9-8)}{8} = \frac{a}{8}, \] so $a$ must be a multiple of $8$. Smallest such $a$ is $8$, giving $b=1$ and \[ S = 3\cdot 8 + 2\cdot 1 + 3 = 29. \] Case 2: $c = 4$ \[ b = \frac{a(12-8)}{8} = \frac{4a}{8} = \frac{a}{2}, \] so $a$ must be even. Take the smallest even $a=2$: \[ b = 1,\quad (a,b,c) = (2,1,4) \text{ are distinct}. \] Then \[ S = 3\cdot 2 + 2\cdot 1 + 4 = 6 + 2 + 4 = 12. \] Case 3: $c = 5$ \[ b = \frac{a(15-8)}{8} = \frac{7a}{8}, \] so $a$ is a multiple of $8$. With $a=8$, $b=7$ and \[ S = 3\cdot 8 + 2\cdot 7 + 5 = 43>12. \] Case 4: $c = 8$ \[ b = \frac{a(24-8)}{8} = 2a. \] With $a=1$, $b=2$ and $(a,b,c)=(1,2,8)$ distinct, \[ S = 3\cdot 1 + 2\cdot 2 + 8 = 15>12. \] Higher values of $c$ only increase $S$, so the minimum value occurs in Case 2 with \[ (a,b,c) = (2,1,4), \quad S_{\min} = 12. \] Therefore, the smallest possible value of \(3a + 2b + c\) is \(12\).