Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 171572 by Raxreedoroid last updated on 18/Jun/22

Show that 4^(3n−2) +5 is divisible by 9 then simplify ((4^(3n−2) +5)/9)

Showthat43n−2+5isdivisibleby9thensimplify43n−2+59

Commented by kaivan.ahmadi last updated on 18/Jun/22

n=1⇒9∣4+5 n=k⇒9∣4^(3k−2) +5 n=k+1⇒4^(3(k+1)−2) +5=4^(3k−2) ×4^3 +5= 4^3 (4^(3k−2) +5)−(4^3 −1)×5= 4^3 (4^(3k−2) +5)−63×5 with hypothesis 9∣4^3 (4^(3k−2) +5) and clearly 9∣63×5 .■

n=1⇒9∣4+5n=k⇒9∣43k−2+5n=k+1⇒43(k+1)−2+5=43k−2×43+5=43(43k−2+5)−(43−1)×5=43(43k−2+5)−63×5withhypothesis9∣43(43k−2+5)andclearly9∣63×5.◼

Answered by floor(10²Eta[1]) last updated on 18/Jun/22

for all n∈Z, 3n−2=3n+1 (1≡−2(mod3)) 3n−2: (...,−2,1,4,7,....) 3n+1: (...,1,4,7,...) 4^(3n−2) +5=4^(3n+1) +5≡64^n .4+5≡4+5≡0(mod9)

foralln∈Z,3n−2=3n+1(1≡−2(mod3))3n−2:(...,−2,1,4,7,....)3n+1:(...,1,4,7,...)43n−2+5=43n+1+5≡64n.4+5≡4+5≡0(mod9)

Terms of Service

Privacy Policy

Contact: info@tinkutara.com