find the last digit of the following expression:4+9^2+4^3+9^4+4^5+9^6+....+9^100

• so the given problem is......
  4+9^2+4^3+9^4+4^5+9^6+....+9^100
  now we can separate this series into two segments as follows....
  ( 4+4^3+4^5+....4^99)+(9^2+9^4+9^6+....+9^100)
  first we will consider the fast segment....
  4 last digit is 4
  4^3 =64 last digit is 4
  4^5=1324 last digit is 4
  last digit of every component in this segment is 4
  so the last digit of this segment of this segment will be the last digit of 4*50(as of there are 50 elements in the First segment(99+1/2))
  that is 200
  now we will consider the second segment....
  9^2 last digit is 1
  9^4 last digit is 1
  9^6 last digit is 1
  last digit of every component in this segment is 1
  so the last digit of this segment of this segment will be the last digit of 1*50(as of there are 50 elements in the First segment(99+1/2))
  that is 50
  so the last digit of the given problem will be 200+50=250------->0
  o the last digit of the given problem will be--->0(Ans)
• 7 years agoHelpfull: Yes(19) No(0)
• The sum goes like 4+81+64+81*81+.........
  That is the sum of units digit goes like 4+1+4+1+..........up to 100 terms.
  We can consider (4+1+4+1) as a set of 4 numbers and the sum is 10. So 100/4 =25. That is 25 more sets are there and the answer therefore is 25*10= 250 (total sum). So last digit is 0 (Answer).
• 7 years agoHelpfull: Yes(5) No(0)
• last digit of 4^1 = 4 ; 4^3 = 4 ; 4^5= 4....
  So, 99 = 1+(n-1)*2 n = 50 i.e. 4+4+4....50 times i.e. 4*50 = 200 (last digit 0)
  Similarly, last digit of 9^2=1 ; 9^4 = 1....
  So, 100 = 2+(n-1)*2 n=50 i.e. 9+9+9+....50 times i.e. 9*50 = 450 (last digit 0)
  So, 200+450 = 650 i.e. last digit is '0'.
• 7 years agoHelpfull: Yes(2) No(3)
• Ans is 0 .
• 7 years agoHelpfull: Yes(1) No(0)
• Ans is 0 .
• 7 years agoHelpfull: Yes(0) No(0)
• use unit digit cycle for digit 4and 9 so u will get last digit as 0.
• 7 years agoHelpfull: Yes(0) No(0)
• There are two series here: series 1 and series 2
  In the series 1:-
  4+4^3+4^5+.....+4^99
  last digit is 0
  In the series 2:-
  9^2+9^4+9^6+.....+9^100
  last digit is also 0 here.
  
• 7 years agoHelpfull: Yes(0) No(0)
• #include
  int isPrime(int k)
  {
  int flag=0;
  if(k==2)
  {
  flag = 0;
  }
  
  for(int j=2; j
• 5 years agoHelpfull: Yes(0) No(0)
• 4^1=4
  4^2=16
  4^3=64
  4^4=256 i.e., the digit at units place can be either 4 or 6
  similarly , for 9 the digit at units place can be either 9 or 1
  so the above sequence for knowing last digit can be written as:
  4+1+4+1+4+1...........4+1=50(4)+1(50)=200+50=250
  hence the last digit is 0
• 5 years agoHelpfull: Yes(0) No(0)