222^222 divided by 7.remainder is

• 222^222 mod 7
  
  (31*7+5)^222 mod 7
  
  5^222 mod 7
  
  (25)^111 mod 7
  
  
  (7*3+4)^111 mod 7
  
  (4)^111 mod 7
  
  (16)^55.(4)^1 mod 7
  
  (7*2+2)^55.4 mod 7
  
  (2)^55.4 mod 7
  
  (8)^18.8 mod 7
  
  (7+1)^18.8 m0d 7
  
  (1)^18.8 mod 7
  
  1*8 mod 7=1
  
  1 (ans)
• 12 years agoHelpfull: Yes(49) No(5)
• 222/7=(31+5/7)..(mixed fraction)
  so its enuf if we check the remainder of(5/7)^222
  remainder of
  (5/7)^0=1
  (5/7)^1=5
  (5/7)^2=4
  (5/7)^3=6
  (5/7)^4=2
  (5/7)^5=3
  (5/7)^6=1... so on
  the cycle repeats for every 6 times
  now divide 222 by 6, the remainder will be 0
  now check the cycle, the remainder for (5/7)^0 i.e 1
  aswer is 1
• 12 years agoHelpfull: Yes(19) No(1)
• using euler theorem
  7 and 222 are coprime to each other
  
  phi(7) = 6
  therefore 222^6 % 7 = 1
  ((222)^6)^37 % 7 = 1
  
  so remainder is 1 when 222^222 is divided by 7.
• 12 years agoHelpfull: Yes(8) No(0)
• srry guys after (4)^222/7 we convert it (64)^74/7
  then (63+1)^74/7
  63/7 reminder is zero and (1)^74/7
  so reminder is 1 its a right ans................
• 12 years agoHelpfull: Yes(7) No(1)
• (2)^2*111=4^111
  ((4^2)^55)*4
  =((16)^55)*4
  first digit of ((16)^55) is 6
  so,6*4=24
  4 is d ans.
• 12 years agoHelpfull: Yes(3) No(15)
• 222^222 divided by 7
  222 mod 7 gives remainder 5
  problem reduces to 5 ^222 mod 7
  now find out cycle
  5 mod 7 gives rem 5
  5^2 mod 7 gives rem 4
  5^3 mod 7 gives rem 6 or -1
  so it gives +1 rem at 5^6
  so cycle here is 6
  now 222 mod 6 gives 0 remainder
  which means last term of seris
  whihc is 1
  
• 11 years agoHelpfull: Yes(3) No(0)
• apply eulers remainder theroem..
  since 222 and 7 are co primes,the totient function f(x)=7(1-1/7)=6
  i.e,rem(222^6/7)=1=rem(222^12/7)=rem(222^18/7)=.......
  since 222 is a multiple of 6
  rem(222^222/7)=1...once refer eulers theroem to understand it better
• 12 years agoHelpfull: Yes(2) No(0)
• (2^222)/7 can be written as ((2^3)^24)/(2^3-1)
  then the remainder will be 1^24=1.
• 12 years agoHelpfull: Yes(2) No(4)
• 2^222 the cyclisity of 2 is 4 dan 4/7 rimender is 1 so answer is 1
• 12 years agoHelpfull: Yes(1) No(2)
• (222^222)%7=1
  how
  step1 222%7=5
  step2 (5^222) % 7
  ((5^3)^74) % 7
  (125^74) % 7 125%7 = -1
  step3 (-1)^74 % 7
  1 (answer)
• 7 years agoHelpfull: Yes(1) No(0)
• answer is 4
• 12 years agoHelpfull: Yes(0) No(9)
• 222^222/7
  (217+5)^222/7
  in 217/7 reminder is zero so we take 5^222/7
  now (25)^111/7 and then convert (21+4)^111/7
  from 21/7 reminder is zero
  now (4)^111/7
  (2^2)111/7
  (2)^222/7
  so reminder is 2
• 12 years agoHelpfull: Yes(0) No(12)
• plz suggest from where to prepare this type of question..if any any e stuff plz send to hargun.15@gmail.com
• 12 years agoHelpfull: Yes(0) No(2)
• Ans. 4
  A B A*B C (A*B)/C
  222 222 49284 7 7040.571429
  
  7*7040=49280
  49284-49280=4
  
  So the remainder is 4.
• 12 years agoHelpfull: Yes(0) No(2)
• 222(base) is divided by 7=31 r5.so we can write as 5^222=25^111. 25 divided by 7=3 r4. so 4^111=16^55 x4. dividing 16 by 7=2 r2. so 2^55 x4 = 32^5 x4. dividing 32 by 7=4 r4. so 4^5 x 4 = 4^6 = 64^2. dividing 64 by 7=9 r1. so 1^2=1
• 11 years agoHelpfull: Yes(0) No(0)
• correct answer is 5.....bcz if u divide 222 by 7 u will get 5 as remainder.....so u convert the given sum as (222)*(222)^221....and divide the single 222 with the 7...u will get remainder as 5
  
• 8 years agoHelpfull: Yes(0) No(0)