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:
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:
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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).
Updated On: Jul 7, 2026
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Correct Answer: 12
Approach Solution - 1
Approach: Solve for the most "expensive" variable so the search is small. Here \(b = \dfrac{a(3c-8)}{8}\), so for \(b\) to be a small natural number you want \(3c-8\) tiny and the factor of \(8\) handled by \(a\). Test \(c=3,4,5,\dots\) and stop early.
Step 1: Rearrange \(3ac = 8(a+b)\): \[ 8b = 3ac - 8a = a(3c-8) \;\Rightarrow\; b = \frac{a(3c-8)}{8}. \] Since \(b\) is a natural number we need \(3c-8>0\), i.e. \(c\ge 3\), and \(a(3c-8)\) divisible by \(8\).
Step 2: The target \(S = 3a+2b+c\) grows with \(a\), so we want \(a\) as small as the divisibility allows. Run through \(c\):
Case \(c=3\): \(b=\dfrac{a(1)}{8}=\dfrac{a}{8}\), so \(a\) must be a multiple of \(8\). Smallest \(a=8\Rightarrow b=1\). Distinct? \((8,1,3)\) yes. \(S = 24+2+3 = 29\).
Case \(c=4\): \(b=\dfrac{a(4)}{8}=\dfrac{a}{2}\), so \(a\) must be even. Smallest \(a=2\Rightarrow b=1\). Distinct? \((2,1,4)\) yes. \(S = 6+2+4 = 12\).
Case \(c=5\): \(b=\dfrac{7a}{8}\), needs \(a=8\Rightarrow b=7\). \(S = 24+14+5 = 43\).
Case \(c=6\): \(b=\dfrac{10a}{8}=\dfrac{5a}{4}\), needs \(a=4\Rightarrow b=5\). \(S=12+10+6=28\).
Step 3: Every larger \(c\) forces either a larger \(a\) (for divisibility) or a larger \(b\), pushing \(S\) up. The smallest value found is in the \(c=4\) case with \((a,b,c)=(2,1,4)\), all distinct naturals, giving \[ S = 3(2)+2(1)+4 = 12. \] Final answer: \(12\).
Step 1: Rearrange \(3ac = 8(a+b)\): \[ 8b = 3ac - 8a = a(3c-8) \;\Rightarrow\; b = \frac{a(3c-8)}{8}. \] Since \(b\) is a natural number we need \(3c-8>0\), i.e. \(c\ge 3\), and \(a(3c-8)\) divisible by \(8\).
Step 2: The target \(S = 3a+2b+c\) grows with \(a\), so we want \(a\) as small as the divisibility allows. Run through \(c\):
Case \(c=3\): \(b=\dfrac{a(1)}{8}=\dfrac{a}{8}\), so \(a\) must be a multiple of \(8\). Smallest \(a=8\Rightarrow b=1\). Distinct? \((8,1,3)\) yes. \(S = 24+2+3 = 29\).
Case \(c=4\): \(b=\dfrac{a(4)}{8}=\dfrac{a}{2}\), so \(a\) must be even. Smallest \(a=2\Rightarrow b=1\). Distinct? \((2,1,4)\) yes. \(S = 6+2+4 = 12\).
Case \(c=5\): \(b=\dfrac{7a}{8}\), needs \(a=8\Rightarrow b=7\). \(S = 24+14+5 = 43\).
Case \(c=6\): \(b=\dfrac{10a}{8}=\dfrac{5a}{4}\), needs \(a=4\Rightarrow b=5\). \(S=12+10+6=28\).
Step 3: Every larger \(c\) forces either a larger \(a\) (for divisibility) or a larger \(b\), pushing \(S\) up. The smallest value found is in the \(c=4\) case with \((a,b,c)=(2,1,4)\), all distinct naturals, giving \[ S = 3(2)+2(1)+4 = 12. \] Final answer: \(12\).
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Approach Solution -2
Method: From \(3ac = 8(a+b)\), solve for \(b\): \(b = \dfrac{a(3c-8)}{8}\). For \(b\) to be a positive integer we need \(c \geq 3\) (so that \(3c-8>0\)) and \(8\) must divide \(a(3c-8)\).
For each \(c\), the smallest valid \(a\) is \(a = \dfrac{8}{\gcd(8,\,3c-8)}\), and since \(3a+2b+c\) increases with \(a\), checking small \(c\) values is enough.
Take \(c=4\): \(3c-8=4\), \(\gcd(8,4)=4\), so the smallest \(a = 8/4 = 2\), giving \(b = \dfrac{2\times4}{8} = 1\). Here \((a,b,c)=(2,1,4)\), all distinct, and \(3a+2b+c = 6+2+4=12\).
Checking nearby cases, \(c=3\) forces \(a=8,\,b=1\) (sum \(29\)); \(c=6\) forces \(a=4,\,b=5\) (sum \(28\)); \(c=8\) gives \(a=1,\,b=2\) (sum \(15\)); every other \(c\) gives a larger sum. So the minimum occurs at \((2,1,4)\).
\[ \boxed{3a+2b+c = 12} \]
For each \(c\), the smallest valid \(a\) is \(a = \dfrac{8}{\gcd(8,\,3c-8)}\), and since \(3a+2b+c\) increases with \(a\), checking small \(c\) values is enough.
Take \(c=4\): \(3c-8=4\), \(\gcd(8,4)=4\), so the smallest \(a = 8/4 = 2\), giving \(b = \dfrac{2\times4}{8} = 1\). Here \((a,b,c)=(2,1,4)\), all distinct, and \(3a+2b+c = 6+2+4=12\).
Checking nearby cases, \(c=3\) forces \(a=8,\,b=1\) (sum \(29\)); \(c=6\) forces \(a=4,\,b=5\) (sum \(28\)); \(c=8\) gives \(a=1,\,b=2\) (sum \(15\)); every other \(c\) gives a larger sum. So the minimum occurs at \((2,1,4)\).
\[ \boxed{3a+2b+c = 12} \]
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