What will be the remainder when 222 222 is divided by 7

(7*31+5)^222 /7

5^222 /7

(5^2)^111 /7
(7*3+4)^111/7
(4^111)/7
(2^3)^74/7
(7+1)^74/7
so, 1/7
i.e remainder=1 ansr sry previous 1 has a little mistake bt this 1 is right
11 years agoHelpfull: Yes(20) No(0)
rem of (222/7)= 5
so now rem of (5^222)/7

now checking cyclicity of rem in 5's power when dvided by 7
rem(5/7)=5
rem(25/7)=4
rem(125/7)=6
rem(625/7)=2
rem(3125/7)=3
rem(15625/7)=1

it repeats after 6 iteration.
so rem(222/6)=0

so rem of (222)^222 is divided by 7 is 1
11 years agoHelpfull: Yes(13) No(3)
(2+2+2)^(2+2+2)
6^6
=46656
now 46656/7=1 reminder
ans is reminder=1
11 years agoHelpfull: Yes(9) No(1)
ans.1 222^222 means unit digits are 2,4,8,6 this digit repeat again &again hence
222/4=55 & means the 3rd digit repetition 8/7=1
11 years agoHelpfull: Yes(3) No(0)
222^222/7= 5^222/7=125^74=(-1)^74/7=1 Ans.
11 years agoHelpfull: Yes(3) No(0)
(17*13+1)^222
(1)^222
anything to the power 1 is 1
so rem is 1...
11 years agoHelpfull: Yes(2) No(2)
remainder w'll be 1
11 years agoHelpfull: Yes(1) No(0)
Ans is 1 ,its very easy, simply apply reminder theorem :)
11 years agoHelpfull: Yes(1) No(4)
firtst divide base by 7 then find remainder then if diviser is prime then substract by 1 and divide power by new diviser
222/7=5 rem and here 7 is prime so new diviser =7-1=6 and 222/6=0 rem
now 5^0/7=1rem
11 years agoHelpfull: Yes(1) No(0)

What will be the remainder when
(222)^222 is divided by 7?