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Because the digit sum of the required number is $15$ ,it must be divisible by $3$ and because it is given that the number is divisible $35$ ,it must be divisible by $5$ and $7\Rightarrow$ the number is divisible by $3,5,7$ .The smallest number satisfying this is $105$ .Looking at the multiples of $105$ ,we get that $735$ is the only three-digit number satisfying the given conditions.P.S:we only have to look at the numbers $105*1,105*2,105*3.........105*9$ because after that it wouldn't be a three-digit number.

Surely, the last digit must be 5.The other two digits must add up to 10. (1,9),(2,8),(3,7),(4,6),(5,5) are possible combinations of first 2 digits.The number 700 is divisible by 35.So, 735 is divisible by 35.

Just continuously add 35 till it become "3 digit" and "sum of digits is 15" :)

The 3 digit number has to be divisible by 3 and 35, but can´t be divisible by 9 once the digit sum is 15 and 15 isn´t divisible by 9. So the number must be a mutiple of 35x3=105. And the number that has the digit sum equals 15 at the same time is multiple of 105 is 735.

I have solved in other way. Given:-A 3-digit number has sum 15 and is divisible by 35. Therefore, ones place can be 5 or 0. But 0 is less possible to happen as its sum is 15.

So let us see if the ones place is 5 . So if the one place is 5 then the sum of other two digits is (15-5)=10. Thus we have find other two digits whose sum is 10 and so that the 3-digit number is divisible by 35. 35=5X7, as the ones place is 5 so it is divisible by 5 thus we have to look for a combination of numbers whose sum is 10 and is divisible by 7. By trial error method we find that 7+3=10 and if we arrange 7 in hundredths place and 3 in tens place and 5 is already arranged in in ones place. The number formed is 735 whose sum is 15 and is divisible by 35.

Three digit no in multiple of 35 is 105,but dig sum is not equal to 15.The no divisible by 35,It must divisible by 5 & 7.so 105 5=525 not true.then 105 7=735 &7+3+5=15.so the ans is 735