Q. Last digit of given expression
(36572)^145 * (30766)^132 will be..

• (36572)^145 * (30766)^132
  last digit of expression will be the last digit of (2)^145 * (6)^132
  
  last digit of 2^145 = (2^4)^36 *2 = (16)^36 *2 = 6*2 = 12 => 2
  
  last digit of 6^132 = 6
  so, Last digit of given expression = 2*6 =12 => 2
  
  ans 2
• 11 years agoHelpfull: Yes(8) No(0)
• 2^145
  2 to power foloow pattern 2^1=2
  2^2=4
  2^3=8
  2^4=6
  .
  .
  2^145 last digit=145/4=1 reminder
  so last digit=2
  6^132 always gives last digit =6
  2*6=12
  ans. 2
• 10 years agoHelpfull: Yes(2) No(0)
• (36572)^145*(30766)^132
  Required digit= unit digit in(2)^145*(6)^132
  (6)^132 always give the unit digit 6.
  (2)^145=(2)^144*2=(2^6)^24*2=(4)^24*2=(64^2)^12*2=(4^2)^12*2=(16)^12*2=(6)^12*2
  =6*2=12=2
  so digit in unit place=2*6=12=>2, so 2 is the answer.
• 11 years agoHelpfull: Yes(1) No(0)
• unit digit of 36572 is 2
  then 2^145 or 2^5 or 32 then unit digit will be 2
  Again unit digit of 30766^132 will be 6.
  Now we have 2*6 or 12 or unit digit will be 2
  Final answer is 2
• 10 years agoHelpfull: Yes(0) No(0)
• hi rakesh
  how (16)^36 *2 =6*2 = 12 => 2 becomes?
  pls explain....
• 10 years agoHelpfull: Yes(0) No(0)
• unit digit of x^y is same as (x mod 10) ^(y mod 4).
  if y is divisible by 4 then we take 4 as power.
  then the problem will be
  2^145 * 6^132
  = 2^1 * 6^ 4 ( since 145mod4 =1 and 132 is divisible by 4)
  =2*1296
  =2592
  therefore unit digit of 2592 = unit digit of (36572)^145 * (30766)^132
  i.e. = 2
  
• 10 years agoHelpfull: Yes(0) No(0)