Question

How many positive integers less than 1000 have the property that the sum of the digits of each such numbers is divisible by 7 and the number itself is divisible by 3

Answer 1

For a number to be divisible by 3, it’s sum of digits must be divisible by 3.

According to question, the sum of digits is divisible by 7 ;

Therefore taking LCM of 7 & 3 we get 21.

The condition will be considered satisfactory only if the sum of digits is 21.

The pair of 3 digits which has sum of 21 are  

{(9,9,3) , (9,8,4) , (9,7,5) , (9,6,6) , (8,8,5) , (8,7,6) , (8,6,5) , (7,7,7) }

(7,7,7) — (777) or 3!/3! = 1 time

(9,9,3) — (399) , (939) , (993) or 3!/2! = 3 times

Similar for (9,6,6) , (8,8,5)

(9,8,4) — (489) , (498) , (849) , (894) , (948) , (984) or 3! = 6 times

Similar for (9,7,5) , (8,7,6) , (8,6,5)  

Total = 1 + 3*3 + 6*4 = 34