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Question Number 126765 by mnjuly1970 last updated on 24/Dec/20

$$...elementarymathematics...$$$$if13\mid 9^{51}+k+1,k\in \mathbb{N}$$$$then{k}_{\left(min\right)}=?$$$$$$$$$$

Answered by floor(10²Eta[1]) last updated on 24/Dec/20

$$9^{51}+\mathrm{k}+1\equiv 0\left(\mathrm{mod}13\right)$$$$9^{51}={\left(9^4\right)}^{12+3}\equiv 9^3\equiv {(-4)}^3=-64\equiv 1\left(\mathrm{mod}13\right)$$$$9^{51}+\mathrm{k}+1\equiv \mathrm{k}+2\equiv 0\left(\mathrm{mod}13\right)$$$$\Rightarrow \mathrm{k}\equiv 11\left(\mathrm{mod}13\right)$$$${\mathrm{k}}_{\min }=11$$

Commented by mnjuly1970 last updated on 24/Dec/20

$$grateful..$$

Commented by JDamian last updated on 25/Dec/20

$${\left(9^4\right)}^{12+3}={\left(9^4\right)}^{15}=9^{60}\ne 9^{51}$$$$9^{51}={\left(9^4\right)}^{12}\times 9^3\ne {\left(9^4\right)}^{12+3}$$

Commented by floor(10²Eta[1]) last updated on 25/Dec/20

$$\mathrm{you}\mathrm{multiply}4.12\mathrm{first}\mathrm{bro}{\mathrm{that}}^{\prime }\mathrm{s}\mathrm{basic}$$$${\left(9^4\right)}^{12+3}\ne {\left(9^4\right)}^{(12+3)}$$$$$$

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