The number of all 3-digit numbers abc in base10 for which a - b - c - a - b - b - c - c - a - a - b - c - 29 is

(a × b × c) + (a × b) + (b × c) + (c × a) + (a + b + c) = 29 (1 + a)(1 + b)(1 + c) = 30 2 × 3 × 5 → (a, b, c) nbsp; ⇒ (1, 2, 3) nbsp;⇒ 6 1 × 6 × 5 → (a, ...

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Standard VII

Mathematics

Multiplication of Integers

The number of...

Question

The number of all 3-digit numbers abc (in base10) 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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Solution

The correct option is C
14

$(a \times b \times c) + (a \times b) + (b \times c) + (c \times a) + (a + b + c) = 29$

$(1 + a) (1 + b) (1 + c) = 30$

$2 \times 3 \times 5 \to (a, b, c) \Rightarrow (1, 2, 3) \Rightarrow 6$

$1 \times 6 \times 5 \to (a, b, c) \Rightarrow (0, 5, 4) \Rightarrow 4$

$1 \times 3 \times 10 \to (a, b, 1) \Rightarrow (0, 2, 9) \Rightarrow 4$

______

14

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The number of all 3-digit numbers abc (in base10) 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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set

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and set

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Then the value of

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A, B, C

such that set A contains all three digit numbers that are multiples of

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set

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contains all three digit even numbers that are multiples of

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and set

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The number of all 3-digit numbers abc (in base10) for which (a×b×c)+(a×b)+(b×c)+(c×a)+a+b+c=29 is

The correct option is C 14 (a×b×c)+(a×b)+(b×c)+(c×a)+(a+b+c)=29 (1+a)(1+b)(1+c)=30 2×3×5→(a,b,c)⇒(1,2,3)⇒6 1×6×5→(a,b,c)⇒(0,5,4)⇒4 1×3×10→(a,b,1)⇒(0,2,9)⇒4 ______ 14

The number of all 3-digit numbers abc (in base10) for which $(a \times b \times c) + (a \times b) + (b \times c) + (c \times a) + a + b + c = 29$ is