find last digits of (1023^3923)+(3081^3921)

• we have (1023^3923)+(3081^3921)
  (3923/4)=4n+3 (3921/4)=4n+1 where n can be any integer
  now through cyclicity we can obtain or you can learn the chart from arun sharma
  number ending in 4n+1 4n+2 4n+3 4n
  1 1 1 1 1
  2 2 4 8 6
  3 3 9 7 1
  
  
  in this way 7+1=8 is the unit's place digit
• 9 years agoHelpfull: Yes(26) No(6)
• A useful tip to solve this type of problems :
  
  power cycle for numbers from 1 to 9
  (1) --- {1}
  (2) --- {2,4,8,6}
  (3) --- {3,9,7,1}
  (4) --- {4,6}
  (5) --- {5}
  (6) --- {6}
  (7) --- {7,9,3,1}
  (8) --- {8,4,2,6}
  (9) --- {9,1}
  
  In this case it's (1023^3923)
  From the above table we can see that the number of elements in power cycle for 3 is 4 i.e {3,9,7,1}
  So divide 3923 by 4 and note the reminder. Here the reminder is 3 and so 7 will be the answer as 7 is present in the third position of power cycle.
  
  Similarly get the last digit of (3081^3921) as 1 and then add both
  1+7=8 is the answer.
• 9 years agoHelpfull: Yes(20) No(1)
• last digit of 1023^3923 = 7
  and last digit of 3081^3921 = 1
  so, 7+1 = 8
• 9 years agoHelpfull: Yes(9) No(2)
• 1023^3923+3081^3921
  dividing both the exponential powers with 4
  therefore, 3923/4=4n+3 and 3921/4=4n+1
  we know,
  4n+1 4n+2 4n+3 4n+4
  for 1 1 1 1 1
  for 2 2 4 6 8
  for 3 3 9 7 1
  and so on...
  therefore, 4n+3(for 3)=7 and 4n+1(for 1)=1
  therefore the last digits will be 7+1 i.e 8
• 9 years agoHelpfull: Yes(7) No(1)
• 3^3+1^1=7+1=8
• 9 years agoHelpfull: Yes(5) No(0)
• 2
  1023^3923 as 3^4 will give 3 in unit place thus 3923 must be a divisible by 4
  thus 1023^3*1023^3920=7*3=21 thus 1 in unit place
  3081^3921=1 in unit place
  thus1+1==2
• 9 years agoHelpfull: Yes(3) No(6)
• 8
  unit digit in 1023^3923=unit digit of 3^3923= unit digit in [(3^4)^980*3^3]
  = unit digit in [1^980] * unit digit
  in [27] = 1 * 7 = 7
  unit digit in 3081^3921=unit digit in 1^3921=1
  therefore unit digit of (1023^3923)+(3081+3921)=1+7=8
  Ans:8
• 9 years agoHelpfull: Yes(3) No(0)
• 1023 take unit digit 3 .cyclicity count of 3 is 4 so now divide 3923 by 4 remainder is 3 ,3 9 7 1 is cyclicity level 3rd pace is 7 so last digit of 1023^3923 is 7.
  for 3081 its unit digit is 1 .1 cyclicity count is 1 so now divide 3921 by 1 remainder will be 0 ,cyclicity level of 1 is 1 only , for remainder 0 we select last level of cyclicity of 1 so unit digit is 1 .
  hence totally 7+1=8
  ans is 8.
• 9 years agoHelpfull: Yes(3) No(1)
• power cycle of 3 is 4.so divide last 2 digits of power value by 4.
  23%4=3.so last digit of first expression is last digit of 3^3=27
  next expression contains last digit as 1.so 7+1=8
• 9 years agoHelpfull: Yes(2) No(0)
• we have (1023^3923)+(3081^3921)
  (3923/4)=4n+3 (3921/4)=4n+1 where n can be any integer
  now through cyclicity we can obtain or you can learn the chart from arun sharma
  number ending in 4n+1 4n+2 4n+3 4n
  1 1 1 1 1
  2 2 4 8 6
  3 3 9 7 1
  
  
  in this way 7*1=7 is the unit's place digit
  in
• 9 years agoHelpfull: Yes(1) No(4)
• using cyclicity of powers.
  (1023^3923) ends with 7 and 3081^3921 ends with 1
  so their sum has 8 in its units place.
  
• 9 years agoHelpfull: Yes(1) No(0)
• we have (1023^3923)+(3081^3921)
  last digit is 3^3+1^1=27+1=28
  last digit is 8
• 9 years agoHelpfull: Yes(1) No(2)
• 0 will be last digit from first we can get 3 as last digit and from last we can get 1 as last digit
• 9 years agoHelpfull: Yes(0) No(5)
• According to power cycle 1023^3923 ends with number 7
  and 3081^3921 ends with number 1
  then 7+1=8.
• 9 years agoHelpfull: Yes(0) No(0)
• 1023 has unit place 3
  so 3 raised to 1 is 3
  3 raised to 2 is 9
  3 to 3 27
  3 to 4 81
  3 to 5 243
  means unit place will change 3 9 7 1 3 9 7 1.......
  so now depend on 3923,there is 4 terms after repition occurs
  so 3923/4=980 and remainder 3 and at 3rd place 7 is there
  now 1 raised to any thing gives 1 at unit place
  when they add up gives 7+1=8
  ans
  
• 9 years agoHelpfull: Yes(0) No(0)
• here 3^3923. 3 is in group c so 3923/4 remainder is 3 so 3^3=unit digit 7
  next 1^anything+init digit is 1
  so (7+1)=8 ans
• 9 years agoHelpfull: Yes(0) No(0)
• power of 3 repeats after 3 times and power 1 is always 1.
  3923=1307*3 + 2
  hence unit place of 1023^3923 = 3^2 =9
  unit place of second term is 1
  thus unit place in problem is 0 (9+1=10)
• 9 years agoHelpfull: Yes(0) No(0)
• last digits means=unit digit,
  now (3^3923)+(1^3921)
  =(3)^(last two digit/4=reminder 3)+(1)
  =3^3 +1=7+1=8
• 9 years agoHelpfull: Yes(0) No(0)
• we have (1023^3923)+(3081^3921)
  (3923/4)=4n+3 (3921/4)=4n+1 where n can be any integer
  now through cyclicity we can obtain or you can learn the chart from arun sharma
  number ending in 4n+1 4n+2 4n+3 4n
  1 1 1 1 1
  2 2 4 8 6
  3 3 9 7 1
  
  
  in this way 7*1=7 is the unit's place digit
  in
• 9 years agoHelpfull: Yes(0) No(2)
• cycle of 3=3,9,7,1. so divide 3923/4. we are left with 3. so 7 is in the 3rd position in the cycle of 3.
  cycle of 1=1
  so ans is 7+1=8
• 9 years agoHelpfull: Yes(0) No(0)