What is sum of last 5 digit in (2020202)^4 ?
Asked on 08 June, 2014

• ans: 17
  
  last 5 digit in (2020202)^4
  can be obtained by last 5 digit in (--20202)^4
  (20202)^4
  2^4*(10101)^4
  16*[(10101)^2]^2
  consider only last 5 digit
  16*( 30201)^2
  16*( 00401)
  06416
  
  sum of last 5 digit in (2020202)^4 = 0+6+4+1+6 = 17
• 10 years agoHelpfull: Yes(163) No(10)
• 16656315024710449921606416
  This is what you get after raising 2020202 to 4 powers.
• 10 years agoHelpfull: Yes(8) No(4)
• last five digits r 06416..... so ans is 17
• 10 years agoHelpfull: Yes(6) No(8)
• I think answer will be ,Ans: 18
• 10 years agoHelpfull: Yes(2) No(22)
• 23 is right if have a doubt then calculate on the calculator...
  
• 10 years agoHelpfull: Yes(2) No(30)
• 16
  is it?
  ..
• 10 years agoHelpfull: Yes(0) No(15)
• I guess 23??
  We don't need all the digits, just last ones would suffice
• 10 years agoHelpfull: Yes(0) No(14)
• 2020202/100000=1010101/50000,remainder is 101.and then 101*2=202=202^4=.....66419 last 5 digit we have to find ,then sum of last 5digit will be 6+6+4+1+9=23
• 10 years agoHelpfull: Yes(0) No(14)
• the approach of the solution given here is wrong..i hv calculated it the last 5digits r 82408 ..thus sum vl b 22
• 10 years agoHelpfull: Yes(0) No(16)
• Ans is 23..
  
  last 3 digit is enough
  
  (202)^4 = 2^4 * 101^4 = 16 * 104060401 = ....66416
  
  so 6+6+4+1+6=23
• 10 years agoHelpfull: Yes(0) No(12)
• frends i am given test on 21-12 but i am not getted result pls tell me what i do..
  
• 10 years agoHelpfull: Yes(0) No(0)
• (2020202)^4=(202)^4
  (202)^4=1664966416
  so last 5 digit=6+1+4+6+6=23
  Ans=23
• 9 years agoHelpfull: Yes(0) No(1)
• Those r saying ans of the question is 23 .they r absolutely wrong .just telling the approach to get easier and earlier solution ...given below
  
  1st take last 5 digit of given no
  Then square of that
  Then take last digit of that result and then multiply with given last 5 digit no
  And continue this until u getting final result.
• 9 years agoHelpfull: Yes(0) No(0)