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CAT 2020 Set-1 | Question: 69

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Given that, $a,b,$ and $c$ are positive integers. That means $a,b,c > 0 \;(\text{or})\; a,b,c \in \mathbb{Z}^{+}$

And, $ab = 432, bc = 96, c<9.$
$$\require{cancel}\begin{array}{|c|c|c|} \hline b \cdot c = 96 & a \cdot b = 432 & a + b + c \\\hline 96 \cdot 1  & - \cdot \cancel{96} & \\ 48 \cdot 2 & 9 \cdot 48 & 9 + 48 + 2  = 56 \\ 32 \cdot 3 & – \cdot \cancel{32} &  \\  24 \cdot 4 & 18 \cdot 24 & 18 + 24 + 4 = 46 \\ 16 \cdot 6 & 27 \cdot 16 & 27 + 16 + 6 = 49 \\ 12 \cdot 8 & 36 \cdot 12 & 36 + 12 + 8 = 56 \\\hline \end{array}$$
$\therefore$ The smallest possible value of $a+b+c = 46.$

Correct Answer$: \text{D}$
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