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Question Number 177621 by aurpeyz last updated on 07/Oct/22

$$\mathrm{For}\mathrm{all}\mathrm{positive}\mathrm{integer}\mathrm{n}.\mathrm{f}\left(\mathrm{n}\right)\mathrm{is}\mathrm{defined}$$$$\mathrm{as}\mathrm{the}\mathrm{remainder}\mathrm{when}\mathrm{n}\mathrm{is}\mathrm{divided}\mathrm{by}\mathrm{k}$$$$\mathrm{where}\mathrm{k}\mathrm{is}\mathrm{an}\mathrm{integer}\mathrm{greater}\mathrm{than}1.$$$$\mathrm{f}(\mathrm{k}+25)=3$$$$\mathrm{Quantity}\mathrm{A}:\mathrm{k}$$$$\mathrm{Quantity}\mathrm{B}:5$$

Commented by aurpeyz last updated on 07/Oct/22

$$plshelp$$

Answered by Rasheed.Sindhi last updated on 08/Oct/22

$$\mathrm{k}+25\equiv 3\left(mod\mathrm{k}\right)$$$$\mathrm{k}+22\equiv 0\left(mod\mathrm{k}\right)$$$$\mathrm{k}+22=\mathrm{uk};\mathrm{u}\in \mathbb{Z}$$$$\mathrm{k}(\mathrm{u}-1)=22$$$$\because \mathrm{k}>1$$$$\therefore Possiblevaluesfor\mathrm{k}:2,11,22$$$$2isnotvalidsinceit{doesn}^{\prime }tsatisfy$$$$\mathrm{f}(\mathrm{k}+25)=3$$$$\therefore \mathrm{k}=11,22$$

Commented by aurpeyz last updated on 09/Oct/22

$$thanks$$

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