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

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

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

Commented by aurpeyz last updated on 07/Oct/22

pls help

$${pls}\:{help} \\ $$

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

k+25≡3(mod k)  k+22≡0(mod k)    k+22=uk;  u∈Z       k(u−1)=22  ∵ k>1    ∴ Possible values for k: 2,11,22  2 is not valid since it doesn′t satisfy  f(k+25)=3  ∴ k=11,22

$$\mathrm{k}+\mathrm{25}\equiv\mathrm{3}\left({mod}\:\mathrm{k}\right) \\ $$$$\mathrm{k}+\mathrm{22}\equiv\mathrm{0}\left({mod}\:\mathrm{k}\right) \\ $$$$\:\:\mathrm{k}+\mathrm{22}=\mathrm{uk};\:\:\mathrm{u}\in\mathbb{Z} \\ $$$$\:\:\:\:\:\mathrm{k}\left(\mathrm{u}−\mathrm{1}\right)=\mathrm{22} \\ $$$$\because\:\mathrm{k}>\mathrm{1}\:\: \\ $$$$\therefore\:{Possible}\:{values}\:{for}\:\mathrm{k}:\:\mathrm{2},\mathrm{11},\mathrm{22} \\ $$$$\mathrm{2}\:{is}\:{not}\:{valid}\:{since}\:{it}\:{doesn}'{t}\:{satisfy} \\ $$$$\mathrm{f}\left(\mathrm{k}+\mathrm{25}\right)=\mathrm{3} \\ $$$$\therefore\:\mathrm{k}=\mathrm{11},\mathrm{22} \\ $$

Commented by aurpeyz last updated on 09/Oct/22

thanks

$${thanks} \\ $$

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