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Question Number 192266 by Shlock last updated on 13/May/23

Answered by Frix last updated on 13/May/23

$$\mathrm{Since}a\in \mathbb{N}\wedge 0⩽a⩽9{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{best}\mathrm{to}\mathrm{solve}\mathrm{for}$$$$n\mathrm{and}\mathrm{try}:$$$$2n^2+14n+83=1111a$$$$n=\frac{-7+\sqrt{2222a-117}}{2}$$$$a=9\wedge n=67\vee a=9\wedge n=-74[\mathrm{if}n\in \mathbb{Z}]$$

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