Q. Find last two digits of 44^70.

• 44^70=> (2*22)^70
  2^70 * 22^70;
  (2^10)^7 * (2*11)^70;
  24*(2^70)*(11^70);
  24*(2^10)^7 * 01;
  24*24*01;
  576;
  so last two digits are 76;
• 10 years agoHelpfull: Yes(1) No(0)
• Cyclicity for 2:
  2^1=2
  2^2=4
  2^3=8
  2^4=6
  2^5=2
  so 2,4,8,6,2,4,8,6...
  Cyclicity for 3:
  3^1=3
  3^2=9
  3^3=7
  3^4=1
  3^5=3
  so 3,9,7,1,3.....
  like these
  for 4: 4,6,4,....
  for 5: 5,5,5,...
  for 6: 6,6,6
  for 7: 7,9,3,1,7,9,...
  for 8: 8,4,2,6
  for 9: 9,1,9,1....
  
  there are four types present
  1)ends in one,
  2)ends in 3,7,9
  3)ends in 4,6,8
  4)ends in 2
  
  @jayakishore lets see ends in one:
  21^786
  let say ab^cde
  so what you have to do is a*e and b^e.
  in order to find the last two digits
  so 2*6=12(take only the last digit i.e. 2)
  and 1^6.(Ans:1)
  so last two digit of 21^786=21
  
  just take another example
  as we know that 11^2=121
  so last two digits 21.
  jst identify this using our method.
  1*2 and 1^2
  so 21.
  
  So, is this understandable for you?
  if you are ok with this then we can see the remaining three types.
• 10 years agoHelpfull: Yes(1) No(0)
• i cant understand how we get 24*(2^70)*(11^70)
• 10 years agoHelpfull: Yes(0) No(0)
• @Jayakishore: we have to convert the numbers in terms of cyclicity..
  Last two digis of (2^10) is 24 so it is raised to odd power.i.e.24.
  and if it is raised to even power means then 76.
  but here odd power i.e.raised to 7... so 24*24*01=576;
  so 76.
  if the given explanation is not enough, let me know.
  
• 10 years agoHelpfull: Yes(0) No(0)