How many of the integers from 1 to 86 (inclusive) contain the digit 4 or have the digit sum divisible by 4?

• numbers having digit 4= { 4,14,24,34,40,41,41,43,44,45,46,47,48,49,54,64,74,84}
  for sum being divisible by 4 sum could be {4,8,12,16) as max sum of 2 digit number is {9+9=18}
  now for 4 ={13,40,31}(digit containing 4 is not included as it is already in set present above )
  for 8 ={17,26,35,53,62,71}
  for 12={39, 57, 66, 75}
  for 16={79}
  total =18+3+6+4+1=32
  so, answer is 32
• 4 years agoHelpfull: Yes(3) No(5)
• 1 to 86
  18= largest possible sum (9+9)
  numbers divisible by 4 < 18 are
  4=>40,4,13,31,22 ==>(5)
  8=>44,08,80,26,62,53,35,71,17 ==>9
  12=>84,48,66,75,57,39 ==>6
  16=>79==>1
  
  digits==>(excluding those considered above for sum condition)
  14,.24,34,41,42,43,45,46,47,49,54,64,74==>13
  
  answer= 13+5+6+9+1=34
• 4 years agoHelpfull: Yes(2) No(0)
• 34 is the answer
  number contain digit 4 is 18 --> 4,14,24,34,40,41,42,43,44,45,46,47,48,49,54,64,74,84.
  for sum being divisible by 4 sum could be {4,8,12,16) as max sum of 2 digit number is {9+9=18}
  now for 4 ={13,22,31}(digit containing 4 is not included as it is already in set present above )
  for 8 ={8,17,26,35,53,62,71,80}
  for 12={39, 57, 66, 75}
  for 16={79}
  
  18+3+8+4+1 = 34
• 4 years agoHelpfull: Yes(1) No(0)
• Numbers ≤86 having digit 4:
  T1=18 Numbers = { 4, 14, 24, 34, 40, 41, 42, 43, 44, 45,46,47, 48, 49, 54, 64, 74, 84}
  Maximum Sum of 2 digits =9+9=18
  
  For the digit sum to be divisible by 4, the sum can be 4,8,12,16
  
  Numbers ≤86 having sum of the digits = 4 and NOT containing 4:
  T2=3 Numbers ={ 13, 22, 31 }
  Numbers ≤86 having sum of the digits = 8 and NOT containing 4:
  T3=8 Numbers ={ 8, 17, 26, 35, 53, 62, 71, 80 }
  Numbers ≤86 having sum of the digits = 12 and NOT containing 4:
  T4=4 Numbers = {39, 57, 66, 75}
  Numbers ≤86 having sum of the digits = 16 and NOT containing 4:
  T5=1 Numbers = { 79 }
  Total possible numbers = T1+T2+T3+T4+T5=18+3+8+4+1=34
  
  Ans: 34
• 4 years agoHelpfull: Yes(1) No(0)
• 25 is the answer
• 4 years agoHelpfull: Yes(0) No(4)
• digits having 4 are 18
  sum divisible by 4 are 14 including 86
  ans is 18+14=32
• 4 years agoHelpfull: Yes(0) No(1)
• 4,24,44,48,64,84
• 4 years agoHelpfull: Yes(0) No(0)
• 4 8 13 14 17 22 24 26 31 34 35 39 40 41 42 43 44 45 46 47 48 49 53 54 57 62 64 66 71 74 75 79 80 84
  total = 34
• 3 years agoHelpfull: Yes(0) No(0)
• Answer: 33
  integers between 1 to 86 contain the digit 4
  
  e.g., 4 , 14, 24, 34 , 40, 41, 42 , 43 , 44 , 45, 46, 47, 48 , 49 , 54 , 64, 74, 84.
  
  total number of integers contain the digit 4 = 18
  
  now we have to find number of integers , have the digit sum divisible by 4.
  
  we know, maximum value of digit sum in case of upto two digit number = (9 + 9) = 18
  
  so, digit sum will be 4, 8, 12, 16 .
  
  case1 :- when digit sum is 4.
  
  possible integers : 13, 22, 31, {excluding integers contain digit 4}
  
  so , number of integers , N1 = 3
  
  case 2 :- when digit sum is 8.
  
  possible integers : 17, 26, 35, 53, 62, 71, 80 {excluding integers contain digit 4}
  
  so, number of integers , N2 = 7
  
  case 3 :- when digit sum is 12.
  
  possible integers : 39, 57, 66, 75 {excluding integers contain digit 4}
  
  so, number of integers = 4
  
  case 4 :- when digit sum is 16,
  
  possible integers : 79
  
  so, number or integer = 1
  
  so, total number of integers = 18 + 3 + 7 + 4 + 1
  
  = 33
• 3 years agoHelpfull: Yes(0) No(0)