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$(0 \leq a < b < c) \in Z$,

$a + b + c + ab + ac + bc + abc = 1622$

$a + b + c = ?$

I've assumed that $a = 0$, by that we can get rid of this part $ab + ac + abc$

Now $bc + b + c = 1622$.

But I found that was useless and got stuck.

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    $\begingroup$ Hint: $(a+1)(b+1)(c+1)$ $\endgroup$
    – vadim123
    Dec 23, 2018 at 16:51

2 Answers 2

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Hint:

$$a+b+c+ab+ac+bc+abc=(1+a)(1+b)(1+c)-1$$

Therefore $(1+a)(1+b)(1+c)=?$

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Note that $$1+a+b+c+ab+bc+ca+abc=(1+a)(1+b)(1+c)$$ $$(1+a)(1+b)(1+c)=1623=1\cdot3\cdot541$$ $$\implies a=0,\ b=2,\ c=540.$$ Note that $0\leq a<b<c$. Those three numbers, 1, 3 and 541 are the only possible factorisation that could satisfy this criteria

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