(201 * 202 * 203 *...............209)^2
last 2 dig=?

• in 205 and 202 we get 5*2=10 and so the lastdigit wil be obviously 0 and 0^2 makes last 2 digits 00
• 10 years agoHelpfull: Yes(15) No(0)
• ANSWER==>00
  using remainder theorem
  (201*202*203.....*209)^2/100= remainder will be last 2 digits
  we get (1*2*3*4*5*6...*9)^2 as remainder
  clearly there are two 5 so last 2 digits= 00
• 10 years agoHelpfull: Yes(4) No(1)
• 0o
  9!=362880^2=00
  
• 10 years agoHelpfull: Yes(1) No(0)
• 01*02*03*......*09=362880
  362880^2 last 2 digits will be 00
  ans is 00
  
• 10 years agoHelpfull: Yes(1) No(0)
• 80
  last numbers are 1,2,3,4......9
  then 1*2*3*......*9=362880
  last two digits =80
• 10 years agoHelpfull: Yes(1) No(5)
• 80
  
  01*02*03*04*....*09=80
  80*80=80
  
  Here I'm just considering the last two digits of each no.
  
• 10 years agoHelpfull: Yes(0) No(2)
• on multiplying unit digit numbers we get 0 and after squaring it we get 00 in last 2 digits.so ans is 00
• 10 years agoHelpfull: Yes(0) No(0)