Q. What is last two digit of 73^435?

• 73^435 = (73^2)^217 * 73
  =(27^2)^217 * 73 {since last two digits of x^2= (100-x)^2 = (50-x)^2 = (50+x)^2}
  =(_29*_29)^217 * 73
  =(_41)^217 *73
  =(_81) * 73
  = _13.
  {since for numbers ending in 1, last digit is always 1 and the second last digit is product of tens digit of base* the unit digit if power, i.e 4*7=28 in this case)
  
• 11 years agoHelpfull: Yes(0) No(0)
• since remainer=435%4=3
  therefore, 73^power=___17
  So last two digit will be 17
• 11 years agoHelpfull: Yes(0) No(0)