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CAT 2025Slot 2QAQues & Sol

Number SystemsHard

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

Solution

1. Concept Used

  • Topic: Number Systems – Optimization under Constraints with Natural Numbers
  • Formula: $3ac = 8(a + b) \implies b = \frac{3ac - 8a}{8} = \frac{a(3c - 8)}{8}$

2. Calculation

We are given the constraint $3ac = 8(a + b)$ where $a, b, c$ are distinct natural numbers, and we want to minimize $3a + 2b + c$.

First, isolate $b$ from the constraint: $b = \frac{3ac - 8a}{8} = \frac{a(3c - 8)}{8}$

For $b$ to be a natural number, $a(3c - 8)$ must be divisible by 8, and we also need $b > 0$, which means $3c > 8$, so $c \geq 3$.

Now substitute $b$ back into the expression to minimize: $3a + 2b + c = 3a + 2 \cdot \frac{a(3c-8)}{8} + c = 3a + \frac{a(3c-8)}{4} + c$

To minimize the objective, we should try small values of $a$ and $c$ while ensuring $a(3c - 8)$ is divisible by 8 and $a, b, c$ are distinct natural numbers.

Case 1: $a = 1$

$b = \frac{1 \cdot (3c - 8)}{8}$

For $b$ to be a natural number, $(3c - 8)$ must be divisible by 8. The smallest such $c$ is $c = 8$ (giving $3(8) - 8 = 16$, so $b = 2$). Check distinctness: $a=1, b=2, c=8$ — all distinct. ✅

$3a + 2b + c = 3(1) + 2(2) + 8 = 3 + 4 + 8 = 15$

Case 2: $a = 2$

$b = \frac{2(3c - 8)}{8} = \frac{3c - 8}{4}$

For $b$ to be a natural number, $(3c - 8)$ must be divisible by 4. The smallest such $c$ is $c = 4$ (giving $3(4) - 8 = 4$, so $b = 1$). Check distinctness: $a=2, b=1, c=4$ — all distinct. ✅

$3a + 2b + c = 3(2) + 2(1) + 4 = 6 + 2 + 4 = 12$

Verification of constraint: $3(2)(4) = 24$ and $8(2 + 1) = 24$ ✅

Case 3: $a = 4$

$b = \frac{4(3c - 8)}{8} = \frac{3c - 8}{2}$

Smallest valid $c$ (odd to make numerator even): $c = 4$ gives $b = 2$, but $a = c = 4$ — not distinct. Try $c = 6$: $b = 5$, so $a=4, b=5, c=6$ — distinct. $3(4)+2(5)+6 = 12+10+6 = 28$ — larger.

Comparing all cases, the minimum value is 12, achieved at $a=2, b=1, c=4$.

3. Solution

Answer = 12 ✅

The smallest possible value of $3a + 2b + c$ is 12, achieved when $a = 2,\ b = 1,\ c = 4$.