2988^678 what is its last three digits?
can anybody give its easy soln

• write 678 in binary format its (1010100110)base2
  the last 3 digits can be obtained by writing the last 3 digits of 988^678
  now 988^1 = 988
  988^2 mod 1000 = 144
  988^4 mod 1000= 144* 144 mod 1000= 736
  988^8 mod 1000 = 736 * 736 mod 1000= 696
  988^16mod 1000 = 696*696 mod 1000=416
  988^32 mod 1000= 416 *416 mod 1000 = 056
  988^64 mod 1000 = 056 * 056 mod 1000= 136
  988^128 mod 1000 = 136*136 mod 1000= 496
  988^256 mod 1000= 496*496 mod 1000= 016
  988^512 mod 1000= 016*016 mod 1000 = 256
  now as we already have 678 in binary form we can write it as 512+128+32+4+2
  988^(512+128+32+4+2) can be calculated as
  (256*496*056*736*144)mod 1000 // using the mod 1000 values calculated above//
  ((256*496)mod 1000*(056*736)mod 1000*144)mod 1000
  (976*216*144)mod 1000
  = 504
  this may not take a long time atleast with a calc
  
  
• 9 years agoHelpfull: Yes(8) No(0)
• Point to be remembered : the last three digits of any multiplication comes by the multiplication of the last three digits of the numbers
  for eg. 1121 * 1154 = 1293634
  if we take the last three digits of the numbers which are :
  
  121 * 154 = 18634
  
  we are getting 634 in both of the results as the last three digits
  
  Answer :
  
  2988^678
  
  we will consider only the last three digits that are 988
  
  so now we will find 988^678
  
  => 988^(2*339)
  we will find the square of 988 i.e. 976144
  but will only consider the last three digits i.e. 144
  we need to find now
  => 144^339
  => 144^(3*113)
  we will find the cube of 144 i.e. 2985984
  Considering only the last three digits, we now have to find
  
  => (984)^113
  => (984^112)*984
  
  using the same concept
  
  => ((984)^(2*56))*984
  
  => (256^56)*984
  
  => (256^(2*28))*984
  
  => (536^28)*984
  
  => (536^(2*14))*984
  
  => (296^14)*984
  
  => (296^(2*7))*984
  
  => (616^7)*984
  
  => (616^6)*(616*984)
  
  considering the last three digits of result of 616*984 i.e. 144
  
  => (616 ^ (2*3))*144
  
  => (456 ^ 3)*144
  
  => (456 ^ 2)*456*144
  
  => 936 * 456 * 144
  
  => 816*144
  
  the last three digits are 504
  
  it is easy but lengthy
• 9 years agoHelpfull: Yes(5) No(4)
• (3000-12)^678
  (12)^678-678*3000*(12)^677+_____________________
  the last three digits are the last 3 digits of 12^678
  
• 9 years agoHelpfull: Yes(1) No(6)
• first calculate 678/4( we get 2 as remainder).now compute 88^2 and we can get 744 as last three digits....so ans is 744.
• 9 years agoHelpfull: Yes(1) No(1)
• (3000-12)^678
  on expansion the terms contain 30000 figure will be atleast three 000 in last
  hence the ans will be 12^678
• 9 years agoHelpfull: Yes(0) No(0)
• i can give the solution upto 2 digits
  
  first divide 678 by 4 and check cyclicity of 8..that will give 4 so the units place will be 4.
  
  now for tens digit take 8(3rd one from lhs) and multiply it with 8(the power's unit digit ) that will give 64 .. so taking its unit place that will be 4..so the last 2 digit will be 44..but i don't have the idea of how to calculate the 3rd one..
  
• 9 years agoHelpfull: Yes(0) No(0)
• ans cannot be 504 or any 144... unit digit of 8^8 is 6...so the unit digit of the ans must be 6.
• 9 years agoHelpfull: Yes(0) No(3)
• 988^2=87144;
  so, Last 3 Digit = 144
• 9 years agoHelpfull: Yes(0) No(0)