The number of all 3-digit numbers abc (in bas | vedclass

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(D) Given the equation: $(abc) + (ab) + (bc) + (ca) + a + b + c = 29$.Adding $1$ to both sides,we get: $(abc) + (ab) + (bc) + (ca) + a + b + c + 1 = 3

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$18$

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(D) Given the equation: $(abc) + (ab) + (bc) + (ca) + a + b + c = 29$.Adding $1$ to both sides,we get: $(abc) + (ab) + (bc) + (ca) + a + b + c + 1 = 30$.This expression factors as: $(a+1)(b+1)(c+1) = 30$.Since $abc$ is a $3$-digit number,$a \in \{1, 2, \dots, 9\}$ and $b, c \in \{0, 1, \dots, 9\}$.Thus,$(a+1) \in \{2, 3, \dots, 10\}$ and $(b+1), (c+1) \in \{1, 2, \dots, 10\}$.We need to find the number of triplets $(a+1, b+1, c+1)$ such that their product is $30$,where $a+1 \geq 2$ and $b+1, c+1 \geq 1$.The prime factorization of $30$ is $2 	imes 3 	imes 5$.The possible sets of factors $(x, y, z)$ where $x = a+1, y = b+1, z = c+1$ are permutations of:$1) (2, 3, 5) ightarrow 3! = 6$ permutations.$2) (1, 3, 10) ightarrow 3! = 6$ permutations.$3) (1, 5, 6) ightarrow 3! = 6$ permutations.$4) (1, 2, 15) ightarrow$ Not possible since $b+1, c+1 \leq 10$.$5) (1, 1, 30) ightarrow$ Not possible since $b+1, c+1 \leq 10$.Total solutions = $6 + 6 + 6 = 18$.

The number of all $3$-digit numbers $abc$ (in base $10$) for which $(a \times b \times c) + (a \times b) + (b \times c) + (c \times a) + a + b + c = 29$ is:

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