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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 abc=a+b+c, how many such (a,b,c)?

$$\mathrm{Given}\:\mathrm{that}\:{a},{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{three}\:\mathrm{consecutive} \\ $$ $$\mathrm{numbers},\:\mathrm{where}\:{a}>{b}>{c},\:\mathrm{such}\:\mathrm{that}\:\mathrm{its}\:\mathrm{product} \\ $$ $$\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{its}\:\mathrm{sum},\:\mathrm{that}\:\mathrm{is}\:{abc}={a}+{b}+{c}, \\ $$ $$\mathrm{how}\:\mathrm{many}\:\mathrm{such}\:\left({a},{b},{c}\right)? \\ $$

Commented byliberty last updated on 06/Nov/20

i think it′s ambiguos. abc= a×b×c or a×100+b×10+c ?

$$\mathrm{i}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{ambiguos}. \\ $$ $$\mathrm{abc}=\:\mathrm{a}×\mathrm{b}×\mathrm{c}\:\:\mathrm{or}\:\mathrm{a}×\mathrm{100}+\mathrm{b}×\mathrm{10}+\mathrm{c}\:? \\ $$

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

$$\mathrm{consecutive}\:\mathrm{numbers}\:\mathrm{means}\:{a},\:{b},\:{c}\:\in\mathbb{Z} \\ $$ $$\Rightarrow\:{b}={a}−\mathrm{1}\wedge{c}={a}−\mathrm{2} \\ $$ $${a}\left({a}−\mathrm{1}\right)\left({a}−\mathrm{2}\right)={a}+\left({a}−\mathrm{1}\right)+\left({a}−\mathrm{2}\right) \\ $$ $${a}\left({a}−\mathrm{1}\right)\left({a}−\mathrm{2}\right)=\mathrm{3}\left({a}−\mathrm{1}\right) \\ $$ $$\left({a}−\mathrm{1}\right)\left({a}^{\mathrm{2}} −\mathrm{2}{a}−\mathrm{3}\right)=\mathrm{0} \\ $$ $$\left({a}−\mathrm{3}\right)\left({a}−\mathrm{1}\right)\left({a}+\mathrm{1}\right)=\mathrm{0} \\ $$ $$\begin{pmatrix}{{a}}\\{{b}}\\{{c}}\end{pmatrix}\:\in\left\{\begin{pmatrix}{−\mathrm{1}}\\{−\mathrm{2}}\\{−\mathrm{3}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\\{−\mathrm{1}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{3}}\\{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}\right\} \\ $$ $$\mathrm{three}\:\mathrm{triples} \\ $$

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