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The number of all 3 -digit numbers abc (in base10) for which $(a \times b \times c)+(a \times b) 6+(c \times a)+a+b+c=29$ is
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The correct answer is:
14
$\begin{array}{l}
(a \times b \times c)+(a \times b)+(c \times a)+(a+b+c)=29 \\
(1+a)(1+b)(1+c)=30 \\
=2 \times 3 \times 5 \rightarrow(a, b, c) \Rightarrow(1,2,3) \Rightarrow 6 \\
=1 \times 6 \times 5 \rightarrow(a, b, c) \Rightarrow(0,5,4) \Rightarrow 4 \\
=1 \times 3 \times 10 \rightarrow(a, b, 1) \Rightarrow(0,2,9) \Rightarrow 4 \\ \quad \frac{14}{} &{ }
\end{array}$
(a \times b \times c)+(a \times b)+(c \times a)+(a+b+c)=29 \\
(1+a)(1+b)(1+c)=30 \\
=2 \times 3 \times 5 \rightarrow(a, b, c) \Rightarrow(1,2,3) \Rightarrow 6 \\
=1 \times 6 \times 5 \rightarrow(a, b, c) \Rightarrow(0,5,4) \Rightarrow 4 \\
=1 \times 3 \times 10 \rightarrow(a, b, 1) \Rightarrow(0,2,9) \Rightarrow 4 \\ \quad \frac{14}{} &{ }
\end{array}$
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