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Question Number 183381 by Shrinava last updated on 25/Dec/22

$$\frac{{3\mathrm{n}}^5+{4\mathrm{n}}^4-{7\mathrm{n}}^3+{5\mathrm{n}}^2-5}{\mathrm{n}+1}$$$$\mathrm{There}\mathrm{can}\mathrm{be}\mathrm{no}\mathrm{residue}:$$$$\mathrm{a})0\mathrm{b})2\mathrm{c})4\mathrm{d})5\mathrm{e})9$$

Commented by Shrinava last updated on 25/Dec/22

$$\mathrm{sorry}:{5\mathrm{n}}^2-4$$

Commented by Frix last updated on 25/Dec/22

$$\mathrm{same}\mathrm{answer}$$

Answered by Frix last updated on 25/Dec/22

$$\frac{3n^5+4n^4-7n^3+5n^2-4}{n+1}=$$$$=3n^4+n^3-8n^2+13n-13+\frac{9}{n+1}$$$$\frac{9}{n+1}=r\iff n=\frac{9}{r}-1\Rightarrow r\ne 0$$

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