16 +17+18+19=70. How many such 4 consecutive numbers are there less than 1000 when added gives a sum which is divisible by 10?

• ans will be 200..
  since 1+2+3+4=10 & 6+7+8+9=30
  any combination with these no be will give u desirable result...
  so total no in 100 will be 20 & that's why in in 1000,it will be 200
  
• 9 years agoHelpfull: Yes(20) No(2)
• Ans will be 200
  
  1. 1+2+3+4=10
  2. 6+7+8+9=30
  3. 11+12+13+14=50
  4. 16+17+18+19=70
  ....
  .....
  .....
  20. 96+97+98+99=390
  
  so from 1 to 100 total nos will be 20
  
  then finally 1 to 1000 nos will be 20*10==200
  
• 9 years agoHelpfull: Yes(13) No(1)
• Ans is : 200
  From 1 to 10 there are 2 pairs (1+2+3+4=_0 & 6+7+8+9=_0)
  So, from 1 to 100 there are 20 pairs
  Hence from 1 to 1000 there will be 20*10=200 such pairs
  
• 9 years agoHelpfull: Yes(5) No(0)
• 1,2,3,4
  6,7,8,9.........996,997,998,999
  
  a+(n-1)d=996
  =>1+(n-1)5=996
  =>n=200
• 9 years agoHelpfull: Yes(4) No(0)
• ans. will be 200.......
  
  
  because four consecutive no. is
  5n+1, 5n+2. 5n+3, 5n+4....
  
  so we can put 0 to 199 no in these four consecutive no...
  
  if we are put 1 in 5n+1 then we will get 6
  
  2 then 7,, 3 then 8 and 4 then 9 so we are add these four consecutive no.. 6+7+8+9=30
  similarly 0 to 199 we can put ... which is less then 1000
• 9 years agoHelpfull: Yes(0) No(0)
• For every 10 there are 8 nos so for every 100 there are 800 nos n for every thousand there are 8000 nos. So total 8000 nos are there for which each 4 consiqutive nos make a no divisible by 10
  
• 9 years agoHelpfull: Yes(0) No(0)
• let us assume that 4 consecutive no. a,a+1,a+2,a+4
  a+a+1+a+2+a+3=10*y
  a=(10*y-6)/4
  a must be divided by 4
  a=1,3,5,7,.......................
  1+(n-1)*2=999
  n=500
  
• 8 years agoHelpfull: Yes(0) No(4)