when sum of the numbers from 9 to 33 divided by 9 the remainder is
a)7
b)5
c)3
d)1

• sum from 1 to 33 =561 i.e. (n(n+1)/2)
  sum from 1 to 8= 36
  561-36= 525
  525/9 remainder is 3.....option C
• 8 years agoHelpfull: Yes(19) No(0)
• s = n/2(a+l)
  = 25/2(9+33)
  = 705
  so remainder is 3
  
  
• 8 years agoHelpfull: Yes(2) No(1)
• 9,10,11,12,13,---------33
  we need to calculate sum of digits of each number 9+1+2+3+4+5+6+7+8+9+10+2+3+4+5+6+7+8+9+10+11+3+4+5+6=147
  147/9 =reminder---->3
  ans:3
• 8 years agoHelpfull: Yes(1) No(0)
• 3 is correct answer.
• 8 years agoHelpfull: Yes(1) No(1)
• Question was: Given sequence 91011....33 .what is the remainder when divided by 9?
  a)2
  b)0
  c)3
  d)7
• 8 years agoHelpfull: Yes(0) No(1)
• Add 9+10+11+....+33. You'll get 525. Divide it by 9, the remainder is 3.
• 8 years agoHelpfull: Yes(0) No(2)