Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 95 by rajabhay last updated on 25/Jan/15

What is the remainder when 17^(23) is divided by 16?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\mathrm{17}^{\mathrm{23}} \:\mathrm{is} \\ $$$$\mathrm{divided}\:\mathrm{by}\:\mathrm{16}? \\ $$

Answered by rajabhay last updated on 03/Dec/14

17^(23) =(16+1)^(23) From binomial theorem (a+b)^n =Σ_(k=0) ^n ^n C_k a^k b^(n−k) (16+1)^(23) =^(23) C_0 16^0 1^(23) +^(23) C_1 16^1 1^(22) +...+^(23) C_(23) 16^(23) All terms other than 1st include power of 16. hence remainder=1

$$\mathrm{17}^{\mathrm{23}} =\left(\mathrm{16}+\mathrm{1}\right)^{\mathrm{23}} \\ $$$$\mathrm{From}\:\mathrm{binomial}\:\mathrm{theorem}\: \\ $$$$\left({a}+{b}\right)^{\mathrm{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\:^{{n}} {C}_{{k}} {a}^{{k}} {b}^{{n}−{k}} \\ $$$$\left(\mathrm{16}+\mathrm{1}\right)^{\mathrm{23}} =\:^{\mathrm{23}} {C}_{\mathrm{0}} \mathrm{16}^{\mathrm{0}} \mathrm{1}^{\mathrm{23}} +\:^{\mathrm{23}} {C}_{\mathrm{1}} \mathrm{16}^{\mathrm{1}} \mathrm{1}^{\mathrm{22}} +...+\:^{\mathrm{23}} {C}_{\mathrm{23}} \mathrm{16}^{\mathrm{23}} \\ $$$$\mathrm{All}\:\mathrm{terms}\:\mathrm{other}\:\mathrm{than}\:\mathrm{1st}\:\mathrm{include}\:\mathrm{power}\:\mathrm{of}\:\mathrm{16}. \\ $$$$\mathrm{hence}\:\mathrm{remainder}=\mathrm{1} \\ $$

Answered by 123456 last updated on 22/Dec/14

17^(23) ≡1^(23) ≡1(mod 16)

$$\mathrm{17}^{\mathrm{23}} \equiv\mathrm{1}^{\mathrm{23}} \equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{16}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com