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Question-192266




Question Number 192266 by Shlock last updated on 13/May/23
Answered by Frix last updated on 13/May/23
Since a∈N∧0≤a≤9 it′s best to solve for  n and try:  2n^2 +14n+83=1111a  n=((−7+(√(2222a−117)))/2)  a=9∧n=67 ∨ a=9∧n=−74 [if n∈Z]
$$\mathrm{Since}\:{a}\in\mathbb{N}\wedge\mathrm{0}\leqslant{a}\leqslant\mathrm{9}\:\mathrm{it}'\mathrm{s}\:\mathrm{best}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{for} \\ $$$${n}\:\mathrm{and}\:\mathrm{try}: \\ $$$$\mathrm{2}{n}^{\mathrm{2}} +\mathrm{14}{n}+\mathrm{83}=\mathrm{1111}{a} \\ $$$${n}=\frac{−\mathrm{7}+\sqrt{\mathrm{2222}{a}−\mathrm{117}}}{\mathrm{2}} \\ $$$${a}=\mathrm{9}\wedge{n}=\mathrm{67}\:\vee\:{a}=\mathrm{9}\wedge{n}=−\mathrm{74}\:\left[\mathrm{if}\:{n}\in\mathbb{Z}\right] \\ $$

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