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Question Number 90379 by jagoll last updated on 23/Apr/20

Commented by john santu last updated on 23/Apr/20

$$x=p,y=p+1,z=p+2$$$$x.z=p^2+2p$$$$\Rightarrow p(p+1)(p+2)=21(3p+3)$$$$(p+1)\{p^2+2p-63\}=0$$$$thenx.z=63$$

Commented by jagoll last updated on 23/Apr/20

$$\mathrm{thank}\mathrm{sir}$$

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