Find the sum of its last five digits (2020202) ^4

• this is equal to 2^4*(1010100+1)^4.
  
  now use binomial.
  
  2^4*[(1010100)^4+4*(1010100)^3+6*(1010100)^2+4*(1010100)+1].
  
  so last 5 digits=16[6*10000+40400+1]=16*00401=06416.
  
  so sum=17
• 9 years agoHelpfull: Yes(36) No(4)
• the answer will be 17
• 9 years agoHelpfull: Yes(3) No(0)
• Right ans is 18 =16*0405
• 9 years agoHelpfull: Yes(1) No(7)
• 7=00016
  Thats it
  
• 9 years agoHelpfull: Yes(0) No(7)
• Right ans is 18 =16*0405
• 9 years agoHelpfull: Yes(0) No(7)
• answer should be 36
  
• 9 years agoHelpfull: Yes(0) No(3)
• answer is 17
  
• 9 years agoHelpfull: Yes(0) No(0)
• Answer is 17 and it is 100% sure
• 9 years agoHelpfull: Yes(0) No(0)
• ans is 23 not 17
• 8 years agoHelpfull: Yes(0) No(0)
• ans is 22 bcoz 2020202^2=last five digits=20804*2020202=last five digits=82408
  so sum is 8+2+4+0+8=22
• 7 years agoHelpfull: Yes(0) No(0)
• let's multiply only last 5 digit (20202)^4=(20001+201)^4
  =((20001+201)^2)^2
  =(00001+40001+2*20001*201)^2) (a+b)^2 considering only last five digit
  =(00001+40001+40402)^2
  =(80404)^2
  = 03216 =12ans; or use (20001+201)^4 through binomial expansion
• 7 years agoHelpfull: Yes(0) No(0)