find out last two digits of (2957^3661)+(3081^3643)
o 42 o 38 o 98 o 22

• For (2957^3661)
  7^1= 7 (unit digit)
  5*1= 5 (tens digit)
  therefore last two digits for this is 57
  For (3081^3643)
  1^3= 1 (unit digit)
  8*3= 24 i.e. 4 (tens digit)
  therefore last two diits for this is 41
  Now 57+41 = 98
  :):)
  
• 12 years agoHelpfull: Yes(47) No(0)
• power cycle:
  take last 2 digits of 1st number i.e 57
  57^1=57
  57^2=49
  57^3=93
  57^4=01
  so last two digits repeat for every 4 multiplications so 3661/4 gives remainder 1, therefore last two digits from above power cycle we get as 57^1=57
  similarly
  81^1=81,81^2=61, 81^3=41, 81^4=21, 81^5=01
  3643/5 gives remainder of 3 so 81^3=41
  last 2 digits of sum=57+41=98
  
• 12 years agoHelpfull: Yes(11) No(1)
• exact procedure to find any no.of digits (can find last 3 digits also)
  any natural number can be written in the form 4n or 4n+1 or 4n+2 or 4n+3
  here
  3661=4(915)+1
  therefore (2957^3661)=(2957^1)=57 here 1 is the remainder from above line
  so last two digits=57
  now 3643=4(910)+3
  therefore (3081^3643)=(3081^3)=====last two digits of (81^3)=41
  therefore 57+41=98
  
  with this procedure we can find any no.of digits like last two,three,fore.....
• 12 years agoHelpfull: Yes(7) No(0)
• answer is 98
• 12 years agoHelpfull: Yes(1) No(3)
• answr is 98
• 12 years agoHelpfull: Yes(1) No(2)
• unit digit of 2957^3661=7^1=7
  then ten digit is 5*1=5
  57
  unit digit of 3081^3643=1^3=1
  tenth digit is 8*3=24
  41
  then solution is 57+41=98
• 12 years agoHelpfull: Yes(0) No(0)
• 98.... power cylces .. 57^1=57...57^4=01(last two digits),57^5=57 =>repeats fr 4 powers.3661 when divided by 4 gives 1 as remainder..hence 57 for 1st part..2nd repeats fr every 5 times..=> 3643 -:5 rem=3 =>41..total 57+41
• 12 years agoHelpfull: Yes(0) No(0)