The units digit of the sum of the first 84 positive integers (that is 1 + 2 + 3 + ... + 84 ) is

• WE KNOW,the sum of a.p series is n/2(a+l)
  here n=84,a=1,l=84
  therefore sum=84/2(1+84)=3570
  so,the unit digit is 0
• 10 years agoHelpfull: Yes(25) No(0)
• sum of first n terms is [n(n+1)]/2
  and here unit digits sum is frm 1to 9....frm 1 to 80
  therefore sum of 1,2,3.......9 is [9(9+1)]/2=45.....
  frm 1 to 80 ...sum is 45*8=360gives 0 at unit place
  now frm 81 to 84...sum of unit digit is:[4(4+1)]/2=20...
  so 360+20 gives 0 at unit place
  so ans is 0...
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• n(n+1)/2=84*85/2;
  
• 10 years agoHelpfull: Yes(1) No(1)
• sum is 84*85/2
  so d last digit is 0.
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• zzzzzzzeeeerrrrrrooooooooooooooooo
  
• 10 years agoHelpfull: Yes(1) No(0)
• sum of 1 to 9 is 45 so the last digit is 5.
  same as for 1 to 80 , the last digit of sum is 8*5=40 i.e. 0
  hence for 81 to 84 the last digit of sum is 1+2+3+4=10 i.e. 0
  hence unit digit of sum is 0
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• 0..
  sum=n(n+1)/2
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• use sum of n natural no formula that is n(n+1)/2. By this we get 0
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• (n*(n+1))/2
  (84*85)/2 = 3570
  unit digit 0
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• 1+2+3+.......+84=84*85/2=3570
  
  Unit digit=0 Ans
  
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