Latest TCS Aptitude Question SOLUTION: How many of the integers from 1 to 86 (inclusive) contain the digit 4 or have the digit sum divisible by 4?

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

Asked In TCS Ggopinath (4 years ago)

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tanamyee parihar  (4 years ago)

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

Is this solution Helpfull? Yes (3) | No (5)  

Radhika Gadve  (4 years ago)

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

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Gnanavinay Murarisetty  (4 years ago)

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

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Loyaally Ranjith  (4 years ago)

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

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Nandita Sahu  (4 years ago)

25 is the answer

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Mary Sheela jangam  (4 years ago)

digits having 4 are 18
sum divisible by 4 are 14 including 86
ans is 18+14=32

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sd  (4 years ago)

4,24,44,48,64,84

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Nishant singh  (3 years ago)

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

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Pavithra Devi K H  (3 years ago)

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

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