Q. 1+2+3+4 which is divisible by 10. Here n is 4. So how many series are possible, starting with 1 such that sum of series is divisible by 10. n should be less than 1000.

• Tthis one is CORRECT
  sum of first n natural no's=n(n+1)/2
  given that, sum should be multiple of 10 and n should less than 1000
  so n can be
  4,24,44,...,984 gives (984-4)/20 +1=50 series
  15,35,55,...,995 gives (995-15)/20 +1=50 series
  19,39,59,...,999 gives (999-19)/20 +1=50 series
  20,40,60,...,980 gives (980-20)/20 +1=49 series
  total no. of series=50+50+50+49=199
  ans 248
• 11 years agoHelpfull: Yes(33) No(5)
• sum of first n natural no's=n(n+1)/2
  given that, sum should be multiple of 10 and n should less than 1000
  so n can be
  4,24,44,...,984 gives (984-4)/20 +1=50 series
  15,25,35,...,995 gives (995-15)/10 +1=99 series
  19,39,59,...,999 gives (999-19)/20 +1=50 series
  20,40,60,...,980 gives (980-20)/20 +1=49 series
  total no. of series=50+99+50+49=248
  ans 248
• 11 years agoHelpfull: Yes(14) No(7)
• sum of n natural no's =n(n+1)/2
  sum should be less than 1000
  so n should be less than 45
  now n(n+1)/2 should multiple of 10 which is satisfied for
  n=4,15,19,20,24,25,35,39,40,44
  so no. of possible series is =10
• 11 years agoHelpfull: Yes(2) No(3)
• n(n+1)/20
  n:- 4,14,24,34............994
  
  994-4/10=99
  99+1=100
  
  100 series possible
• 11 years agoHelpfull: Yes(1) No(7)
• ****
  n(n+1)
  should be a multiple of 20****
  
  case1: n or n+1 is a multiple of 20
  
  possibilities: 19,39,59...999 , 20,40,60....980 (n
• 11 years agoHelpfull: Yes(1) No(3)
• Sn=n(n+1)/2
  Therefore, n(n+1)/2 must be divisible by 10
  Or, n(n+1) must be divisible by 20
  
  CASE 1: take the following consecutive numbers
  4,5,6 [4*5 divisible by 20]
  14,15,16 [15*16 divisible by 20]
  24,25,26 [24*25 divisible by 20]
  34,35,36 [35*36 divisible by 20]
  44,45,46 [44*45 divisible by 20]
  .
  .
  .
  984,985,986 [984*985 divisible by 20]
  994,995,996 [995*996 divisible by 20]
  From above steps it is clear that there is
  an alternative number around the 5's multiple
  which is producing the combination to
  produce a number divisible by 20.
  COUNT_1=100
  
  CASE 2: take the following consecutive numbers
  19,20,21 [19*20 and 20*21 both are divisible by 20]
  39,40,41 [39*40 and 40*41 both are divisible by 20]
  59,60,61 [59*60 and 60*61 both are divisible by 20]
  79,80,81 [79*80 and 80*81 both are divisible by 20]
  .
  .
  .
  979,980,981 [979*980 and 980*981 both are divisible by 20]
  From above steps it is clear that both the numbers
  around the 20's multiple are producing the combination to produce a number divisible by 20.
  COUNT_2= 49*2=98
  
  CASE 3:
  The series 1 to 999 is divisible by 20
  COUNT_3=1
  
  Total Count = Count_1 + Count_2 +Count_3 = 100+98+1 = 199
• 11 years agoHelpfull: Yes(1) No(1)
• ****
  n(n+1)
  should be a multiple of 20****
  
  case1: n or n+1 is a multiple of 20
  possibilities: 19,39,59...999 , 20,40,60....980 (n
• 11 years agoHelpfull: Yes(0) No(2)
• ans is 50+50+50+49 = 199
• 11 years agoHelpfull: Yes(0) No(2)