what is the last two digit of (3243^8897)+(6574^2314)

• 19 is the ans. power cycle of 43(43,49,07,01) is 4 . so 8897/4 will give remainder as 1 then 43 is taken .For 74 the power cycle is 2(76,24) so 2314/2 gives remainder 0 so 76 is taken 76+43=119 last two digits are 19
  
• 11 years agoHelpfull: Yes(22) No(4)
• 3243 is raised to odd power hence the last diggit must be 3 and 6574 is raised to even power its unots digyit must be 6. For tens digit calculate power 4 of 43 and 74^2. And add them. Ans 19
• 11 years agoHelpfull: Yes(5) No(4)
• 19 is the answer.
• 11 years agoHelpfull: Yes(4) No(0)
• 8897 is power of 3243... so take th eunit digit of base.
  (1)3 power 1=3
  (2)3 power 2=9
  (3)3 power 3=27
  (4)3 power 4=81.......
  here unit digit are repeated... so divide power 8897 by 4.. remainder is 1.
  so 3.
  the process for 6574 power of 2314... same do for 4 unit digit and divide the power 2314 by 2.so 16... then add 3+16=19. 19 is the answer
• 11 years agoHelpfull: Yes(3) No(3)
• can any 1 explain it .. i know for last digit but not for last 2 digit yr
• 11 years agoHelpfull: Yes(2) No(2)
• please explain clearly
• 11 years agoHelpfull: Yes(2) No(0)
• ans: 19 is the last 2 digit
• 11 years agoHelpfull: Yes(1) No(0)
• 12 is the correct answer
• 9 years agoHelpfull: Yes(1) No(0)
• 43+69 whose last two digit is 12.
  Ans is 12.
• 11 years agoHelpfull: Yes(0) No(14)
• 3243^8897=83
  6574^2314=86
  ans=69
• 11 years agoHelpfull: Yes(0) No(9)
• M*N=7. 7/6=1.
  
• 11 years agoHelpfull: Yes(0) No(0)
• The last two digits are 19 Ans.
• 11 years agoHelpfull: Yes(0) No(0)