a series of natural number 1+2+3+4+5+6+7+8+9+10 is divisibel by 11 so we has to find how many such series is possible upto n where n is less than 1000.
a.90
b.99
c.180
d 96

• the total multiple of 11 less then 1000 are 90 .
  and for every multiple there are two such series which are divisible by 11.
  . .
  . sum of series = n(n+1)/2
  for multiple 11 we can take n=10 and 11
  for 22 '' '' ''' n=21 and 22
  and so on and for 990 multiple of 11 we can take n=989 and 990
  so total series formed are 90*2= 180
• 8 years agoHelpfull: Yes(23) No(4)
• 99 is the answer
  sum of n natural numbers is n(n+1)/2 for n(n+1)/2 is divisible by 11 then n(n+1) should be divisible by 22 because 2 and 11 co-primes now find out possible n
  n can take 22,44,66,88,....,990 i.e., 22 multiples less than 1000 there are 45 such numbers less than 1000
  n can also take 21,43,65....989 i.e., 1 less than 22 multiples again there are 45 such numbers less than 1000
  we need to look other possibilities as n(n+1) should be divisible by 22 then n(n+1) should be divisible by 2 and 11 so find whether there are consecutive numbers whose product is divisible by 11 and 2 they are
  10*11
  11*12
  121*122
  231*232
  452*452
  561*562
  671*672
  781*782
  891*892
  those are obtained by multiples of 11 which end with 1 bcz next number is even which is divisible by 2 so 11,121,231,451,561,671,781,891 are multiples of 11 so hence there are 9(above mentioned) possibilities
  Total=45+45+9=99
• 8 years agoHelpfull: Yes(3) No(4)
• ans is 99 they are asking how many such series is possible
  if we go through 11 to 99 then 9 such series are possible 10(1+2+...9),20(1+2+3..) already given series is divisible by 11..from 100 to 999 there is 90 such series possible for ex 100(1+2+3+4+5..) for 110(1+2+3+4.....)....990(1+2+3...+9)
  total 90+9 =99 such series are possible
  if i am wrong plzz correct me
  
• 8 years agoHelpfull: Yes(2) No(0)
• please admin gives us correct ans
• 8 years agoHelpfull: Yes(2) No(0)
• 180 is the correct answer.
• 8 years agoHelpfull: Yes(2) No(0)
• Sum of the series=n(n+1)/2
  for being it divisible by 11 , there should be n= multiple of 11 or (n+1)= multiple of 11
  
  for 1st condition n can be:
  11,22,33,..........,990 ( a+ (n-1)d= 990 ,a=11, d=11 => n=90
  
  for 2nd condition (n+1) can be:
  10,21,32,...........,989 ( a+ (n-1)d= 989 ,a=10, d=11 => n=90
• 7 years agoHelpfull: Yes(2) No(0)
• @MALLIKARJUNA REDDY .Why did u take 2 as a co-prime of 11? There are also others co-prime of 11 like 3,4....
  And how did u calculated how many nos are there b/w 22 and 990?
• 8 years agoHelpfull: Yes(0) No(0)
• sorry the ans I think is 180 bcz all multiples of 11 for n is satisfied 11
• 8 years agoHelpfull: Yes(0) No(0)