222 ^222 divided by 7 wt is remainder

• 222=31*7+5
  31*1+5 when divided by 7 gives 5 as remainder
  
  =>5^222
  now 5^2^111
  =>25^111
  (7*3+4)^111 when devided by 7 gives 4 as remainder
  
  =>4^2^55*4^1=16^55*4^1
  =>(7*2+2)^55 when devided by 7 gives 2 as remainder
  =>2^55*4=2^54*2*4=2^54*8=8^18*8
  
  hence 8^18*8=(7+1)^18*8 when divided by 7 gives 1^18*8
  =>8
  
  and when 8 is divided by 7 gives 1 as remainder.
  
  Hence answer will be 1.
  
  Correct me if I am Wrong
  
• 10 years agoHelpfull: Yes(40) No(1)
• 222^222/7
  =(5)^222/7
  =(5^6)^37/7
  (1)^37/7
  =1
  Ans remainder=1.
  
• 10 years agoHelpfull: Yes(21) No(3)
• 222/7=5
  now our question turns to-5^222/7
  now we have to find cyclicity of 5
  5^1/7=5(remainder)
  5^2/7=4(remainder)
  5^3/7=6
  5^4/7=3
  5^5/7=1(remainder)
  cyclicity is 6
  divide power 222 by 6
  remainder 0
  means unit digit is 1(last line of cycle of 5)
  
• 10 years agoHelpfull: Yes(10) No(5)
• 
  222^1/7=5
  222^2/7=4 => (take above remainder)- (5*222)/7=4
  222^3/7=6 => (take above remainder)- (4*222)/7=6
  222^4/7=2 => (take above remainder)- (6*222)/7=2
  222^5/7=3 => (take above remainder)- (2*222)/7=3
  222^6/7=1 => (take above remainder)- (3*222)/7=1
  222^7/7=5 => (take above remainder)- (1*222)/7=5
  5,4,6,2,3,1 is the remainder value
  hence 222^222 divided by 7 has cyclicity=6 because repitition occurs after sixth power.
  therefore,222/6 gives 0 remainder .
  So,we check at 222^6/7 remainder which is equal to 1
  ans. 1
• 10 years agoHelpfull: Yes(4) No(0)
• For 222^222, the unit digit must be 4,if we go through the multiples of 7 then we will find 7*9=63 Whose unit digit is as close as the 4 like no other.so remainder is 1.
• 10 years agoHelpfull: Yes(2) No(1)
• 222/7=5
  (5*5)/7=4
  (4*4)/7=2
  (2*2*2)/7=1
  no more remainder for proceed.
  1 is ans.
• 10 years agoHelpfull: Yes(1) No(1)
• 222^222/7
  ->5^222/7
  ->(5^3)^74/7
  ->6^74/7
  ->36^37/7
  1^37/7
  ->ans is 1...
  
• 10 years agoHelpfull: Yes(1) No(0)
• 4
  222 leaves 5
  5^5=25
  remainder 4
  
• 10 years agoHelpfull: Yes(0) No(2)
• Answer is 121112227as 1117 leave remainder -1 and as power is even thus it become 1so2-1222722227now as 238so237478747Now 87 leave remainder 1 soanswer is 1
• 10 years agoHelpfull: Yes(0) No(1)
• 222^222/7
  firstly divide 222 with 7 and we get 5 remainder now
  (217+5)^222/7 now 217 will give 0 as remainder so left with 5 only so now we have to solve 5^222/7 now take square of 5 so ((5^2)^111)/7 it will make 25^111/7 so this equation will give 4 as remainder
  now take cube of 4 which is 64 1 more than 63 so equation will become (4^3)37/7
  64^37/7
  (63+1)^37/7
  so we get 1 as remainder
• 10 years agoHelpfull: Yes(0) No(0)
• (7*31+5)^222/7
  (5^2)^111/7 => (7*4+1)^111/7
  Hence Remainder is 1
  
• 10 years agoHelpfull: Yes(0) No(0)
• (222)^222/7 = (224 - 2)^222/7
  => (-2)^222/7 = (2)^222/7
  => (2^3)^37 x 2^111/7
  => (8)^37 x 2^111/7 = (7+1)^37 x 2^111/7
  => 1^37 x 2^111/7 = 2^111/7
  => (2^3)^37/7 = (7+1)^37/7
  => 1^37/7 = 1 remainder Ans...
• 10 years agoHelpfull: Yes(0) No(0)
• ((2^3)^74)/7
  =((7+1)^74)/7
  =(1^74)/7
  =1/7
  =1
• 10 years agoHelpfull: Yes(0) No(0)
• 222^222=(222^6)^37%7=1^37%7=7
  as 7 is a prime no, so 222^(7-1)/7=1
• 9 years agoHelpfull: Yes(0) No(0)
• 7 is a prime number so, divided by 6(7-1)
  (222^(222%6))
  ((222^0)%7)=0 so, result=0
  
  
• 9 years agoHelpfull: Yes(0) No(2)