Given that a,b and c are three consecutive numbers, where a>b>c, such that its product is the same as its sum, that is abc=a+b+c, how many such (a,b,c)?


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Question Number 121219 by ZiYangLee last updated on 06/Nov/20

$Given that a, b and c are three consecutive$ $numbers, where a > b > c, such that its product$ $is the same as its sum, that is a b c = a + b + c,$ $how many such (a, b, c) ?$
Commented byliberty last updated on 06/Nov/20

$i think {it}^{'} s ambiguos .$ $abc = a \times b \times c or a \times 100 + b \times 10 + c ?$
	Answered by MJS_new last updated on 06/Nov/20
	$consecutive numbers means a, b, c ∈Z ⇒ b=a−1∧c=a−2 a(a−1)(a−2)=a+(a−1)+(a−2) a(a−1)(a−2)=3(a−1) (a−1)(a^2 −2a−3)=0 (a−3)(a−1)(a+1)=0 ((a),(b),(c) ) ∈{ (((−1)),((−2)),((−3)) ) , ((1),(0),((−1)) ) , ((3),(2),(1) )} three triples$
	$consecutive numbers means a, b, c \in Z$ $\Rightarrow b = a - 1 \land c = a - 2$ $a (a - 1) (a - 2) = a + (a - 1) + (a - 2)$ $a (a - 1) (a - 2) = 3 (a - 1)$ $(a - 1) (a^{2} - 2 a - 3) = 0$ $(a - 3) (a - 1) (a + 1) = 0$ $(\begin{matrix} a \\ b \\ c \end{matrix}) \in {(\begin{matrix} - 1 \\ - 2 \\ - 3 \end{matrix}), (\begin{matrix} 1 \\ 0 \\ - 1 \end{matrix}), (\begin{matrix} 3 \\ 2 \\ 1 \end{matrix})}$ $three triples$
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