Question

The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is____________
(Please give proper working for your answer)

Answer 1

Given : number of numbers, lying between 100 and 1000 formed with the digits 1, 2, 3, 4, 5

the repetition of digits is not allowed

To Find : The total numbers divisible by either 3 or 5

Solution:

numbers, lying between 100 and 1000

Hence 3 digit numbers from 101  to 999

Digits available    1, 2, 3, 4, 5

divisible by  5  iff last digit is 0 or 5

divisible by  5 Hence last digit must be 5    ( as 0 is not here)

divisible by 3 sum of digits must be divisible  by 3

Divisible by 5

last digit 5

remaining 2 digits can be  in ⁴P₂ = 12 Way

Hence 12 numbers

Divisible by 3

123  ,  132  , 135  , 153  

213 ,  231 ,  234  , 243 ,

315 ,  324  , 342  , 345 , 351 , 354

423  , 432  , 435 , 453

513  , 531  , 534 ,  543

18 Numbers

divisible by 3 and 5 both

XY5  ( let say 3 digit number)

X  = 1   => Y  = 3   is only possible ( as 5 +  1= 6   and  6 + 3 = 9)

X = 2  =>   Y  not possible    

X = 3  =>  Y = 1 , 4    possible

X = 4    => y  = 3

4 Such possible numbers only

135 ,  315  , 345  , 435

Hence  18 + 12 - 4  = 26 numbers  divisible by either 3 or 5

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