Dividing By The Digit Sum

some the obvious answers are multiples of 9 which are, 18,27,36,45,54,63,72,81,90

other are that of 10 i.e.

10,20,30,40,50,60,70,80

(not including 90 bcoz it is already included)

keeping in mind that all multiples of three are such that there digit sum is 3 or its multiple

so we have to include all the numbers which are multiples of three and their digit sum is 3

so that are only 12 and 21

(30 is also there but it has been included)

now similarly we observe for multiples of three which are 6,12 .....

but the case is true only for 6 and 12

so we have to find multiples of 6 and 12 that sum 6 and 12 respactively.

that are (for 6) 24 and 42

and (for 12 ) 48 and 84

Now the question is why only 6 and 12????

so the answer is that for a two digit number the highest possible sum is 18 (9+9)

so we reject all multiples of three greater than 18 and 18 itself!!! it is bcoz the only way to get sum 18 is by the number 99

and clearly 18 doesn't divide 99

so the only choice left is 15.

now see that to get digit sum 15 the numbers are 69,78,87,96 and 15 doesn't divide any of them.

so there are total 23 numbers such that they are multiples of their digits.

PLZ GUIDE ME IF I AM WRONG SOMEWHERE.