How many possible combination of four consecutive numbers between 1to 1000 whose sum is divisible by 10.ex;-16+17+18+19=70

• it is possible for all the numbers having unit digit as
  (1,2,3,4) and (6,7,8,9)
  so, b/n 1 to 10 difference there are two possibility (i,e, 1-10,11-20,21-30,etc)
  so total possibility
  = 1000*2/10
  = 200
  200 is the answer
• 10 years agoHelpfull: Yes(60) No(21)
• ans is 199
  2 possibles for every 10
  20 possibles for every 100
  200 possibles for 1000
  but we will not consider 1(1,2,3,4 this not possible)
  so 200-1=199
  
• 10 years agoHelpfull: Yes(43) No(10)
• 11+12+13+14=50 which is divisible by 10 and
  16+17+18+19=70 which is divisible by 10.
  So for 10 to 20 there is two possibilities, hence for 20 to 30 there will be two possibilities .
  so in order to count how many possibilities for 10 to 1000, let
  10,20,30,.....1000, we want to find no. of terms
  last term=a+(n-1)d
  1000=10+(n-1)10
  =>100=1+n-1
  => n=100
  so there will be 100*2 chances
  =200
  
• 10 years agoHelpfull: Yes(15) No(2)
• ans-198 bcz 1 & 1000 not include it is given in question b/w 1 to 1000.
  
• 10 years agoHelpfull: Yes(3) No(10)
• let numbers be x,x+1,x+2,x+3
  sum=4x+6=2(2x+3)
  now for x=1,6,11.........996 it gives the multiple of 10.
  so total no. is 996=1+(n-1)5
  n=200
• 9 years agoHelpfull: Yes(3) No(0)
• ans is 199
  2 possibles for every 10
  20 possibles for every 100
  200 possibles for 1000
  but we will not consider 1(1,2,3,4 this not possible)
  so 200-1=199
  becoz between we will not consider
  
• 9 years agoHelpfull: Yes(2) No(1)
• ans is 33
  
• 10 years agoHelpfull: Yes(1) No(11)
• ans is 200
  
  11+12+13+14=50 ( divisible by 10 )
  16+17+18+19=70 ( divisible by 10)
  from 10 to 20 there is two possibilities
  for 20 to 30 there will be two possibilities .
  so in order to count how many possibilities for 10 to 1000, let us assume that the nos. are in AP
  
  10,20,30,.....1000
  l=a+(n-1)d
  1000=10+(n-1)10
  =>100=1+n-1
  => n=100
  so 100*2 combinations are possible.
  
  
• 10 years agoHelpfull: Yes(1) No(1)
• there are total of 200 combination of number between 1 to 1000
  since in each 1 to 10 , 11 to 20 , 21 to 30 ...............991 to 1000 there are only two combination of four consecutive numbers whose sum is divisible by 10.
• 10 years agoHelpfull: Yes(0) No(0)
• four consecutive numbers : a+b+c+d=10k;
  
  1+2+3+4=10;
  6+7+8+9=30;
  11+12+13+14=50;
  
  now give a look ,here among the possible combination there is pattern followed
  1+5=6
  2+5=7
  3+5=8
  4+5=9
  similarly all combinations follow same pattern.
  now find last combination consecutive digit :
  will come from 1+(n-1)*5;
  take n=200;
  then 1+(200-1)*5=996;
  so last consecutive combination :996+997+998+999=3990
  
  so total number of such combinations are: (996-1/5)+1=200
• 9 years agoHelpfull: Yes(0) No(0)
• 1 to 10 = (1+2+3+4 , 6+7+8+9)=2 combinations
  2 to 20=(11+12+13+14 , 16+17+18+19)=2 combinations
  so total combinations between 1 and 100=10*2=20
  like this total combinations 1 to 1000=20*2=200
• 9 years agoHelpfull: Yes(0) No(1)
• (x+(x+1)+(x+2)+(x+3))/10=(4x+6)/10=(2x+3)/5
  (2x+3)/5=399
  x=996
  
• 9 years agoHelpfull: Yes(0) No(1)
• 10-2p,
  1000-200
  ans-200-1=199
• 9 years agoHelpfull: Yes(0) No(0)
• 1+2+3+4=10
  6+7+8+9=20
  
  11+12+13+14=50
  16+17+18+19=60
  as show on
  for ever 10 (like 1-10,11-20,21-30....)number we get 2 set of consecutive number
  so,
  possible combination is=(1000/10)*2=200
• 9 years agoHelpfull: Yes(0) No(0)