Digit Sum Condition

Clearly, the digit sum of $N+3$ will be greater than that of $N$ if the last digit of $N$ is smaller than 7. Thus the last digit is 7, 8, or 9. Now suppose that $N = \overline{abc}$ , so that the digit sum of $N$ is $a+b+c$ .

Case 1: $b<9$ . The digit sum of $N+3$ will be $a+(b+1)+(c-7) = a+b+c-6$ . Bringing in the original condition gives $a+b+c = 3(a+b+c-6)$ and doing some simple manipulation gives $a+b+c = 9$ . From here, the 3 possibilities of $N$ can be obtained as 108, 117 and 207, and it can be checked easily that these 3 numbers fulfil the given requirements.

Case 2: $b=9$ . Using a similar method, we get $a+b+c=3((a+1)+(b-9)+(c-7))$ , giving $2(a+b+c)=15$ . This is impossible since $a, b$ and $c$ are non-negative integers smaller than 10.

Thus the only solutions are 108, 117 and 207, and the required answer is $108+117+207=432$

[Latex edits - Calvin]

When there's an additional carry, the last digit become 0 from 9, and the digit before it is added 1, so the sum of the digits will decrease by 9 with the exception of the 1 which is added. Let the 3-digit number be abc, and the sum of the digits be x. Then consider four possible cases when 3 is added: If no carry exists, the sum of all its digits will become x+3, there's no solution to the equation x=3 (x+3). If there's one carry, the sum of all its digits become x+3-9=x-6, and the equation x=3 (x-6) has the solution x=9. Because there's only one carry, so c must be no less than 7, and b must be less than 9. Thus, there are three numbers satisfying: 207, 117, 108. If there are two carries, the sum of all its digits become x+3-9 2=x-15, and the equation x=3 (x-15) does not have an integer root. If there are three or more carries, the root of the equation x=3 (x+3-9 n), in which n is no less than 3, is greater than 27, the greatest sum of all digits in a 3-digit number. Thus, the answer is 207+117+108=432.

define x'DS=digit sum of x in the following. let N=abc,first we can infer that c≥7 or digit sum of N+3 can't be smaller than N's digit sum. then let b=9,N'DS=a+9+c=3 (N+3)'DS=3 (a+1+0+c-7),(as a=9 is impossible) then we have 27=2 (a+c) which is also impossible . Now,let c=7,from above we know b<=9 so N'DS=a+b+7=3 (N+3)'DS=3*(a+b+1),get equation a+b=2, so 207 and 117 is legal. Now c=8,get equation a+b=1, so 108 is legal when c=9 ,it is a+b=0 which has no legal soultion.

so answer is 108+207+117.

Let S(N) be the digit sum of N. Since S(N) > S(N+3), then the units digits of N must be 7, 8 or 9. Otherwise, S(N) < S(N+3) since the units digits will increase while the tens and hundreds digits are unchanged. We consider two cases: Case 1: When the tens digit is 0, 1, 2, ..., or 8.This is the case when adding 3 to N doesn't change the hundreds digit of N. Then S(N+3) = S(N) - p + (p+3 - 10) + 1 = S(N) - 6, where p is an element of {7, 8, 9}. Then using the fact that S(N) = 3S(N+3), we get S(N) = 9. Hence N has possible vales 117, 207 and 108. Case 2: When the tens digits is 9. This is the case when adding 3 to N will change the hundreds digit of N. Then S(N+3) = S(N) - p + (p + 3 -10) - 9 + 1 = S(N) - 15. Also using the fact that S(N) = 3S(N+3), we get S(N) = 22.5, which has no solution for N. Hence the only solution for N are 117, 207 and 108 yielding the sum of 432.

Let $N = \overline {abc} = 100a + 10b + c$ , where $1\leq a \leq 9$ , $0 \leq b \leq 9$ and $0 \leq c \leq 9$ .

If $c \leq 6$ , then the digit sum of $N+3$ can only increase. Hence $c\geq 7$ .

If $b=9$ , then by definition $a+9+c =3 \left( (a+1) + 0 + (c+3-10) \right) \Rightarrow 2 a + 2 c = 27$ which has no integer solutions due to parity . Hence $b \leq 8$ .

We now solve $a + b + c = 3 \left[ a + (b+1) + (c+3 - 10) \right] \Rightarrow a + b + c = 9$ .

With the above restrictions, the only solutions are $117, 207$ and $108$ . Therefore the sum is $117 + 207 + 108 = 432$ .