There are exactly twenty-one 3-digit numbers for which the sum of the digits is six: 105, 150, 501, 510, 114, 141, 411, 123, 132, 213, 231, 312, 321, 204, 240, 402, 420, 222, 303, 330, 600

Let the $3$ -digit number be $abc$ where $1 \le a \le 6, 0 \le b \le 6, 0 \le c \le 6$ .

Then the number of suitable $3$ -digit integers is equivalent to the number of integer solutions to the equation $a + b + c = 6$ with the aforementioned restrictions.

Now let $a = a' + 1$ . Then the equation becomes $a' + b + c = 5$ where each of $a',b,c$ is a non-negative integer. This is now a stars and bars problem, (see Theorem $2$ in the link), with solution

$\dbinom{7}{5} = \boxed{21}$ .

Comments: Using the method discussed above, to find the number of suitable $2$ -digit numbers, we would be working with the equation $a' + b = 5$ , giving a solution of $\binom{6}{5} = 6$ , as expected.

For $4$ -digit numbers, we would have $a' + b + c + d = 5 \Longrightarrow \binom{8}{5} = 56$ numbers.

In general, for an $k$ -digit number there would be $\binom{k+4}{5}$ solutions. So, for example, there would be $\binom{14}{5} = 2002$ $10$ -digit numbers whose digits sum to $6$ .

• If the hundred digit is 1, there are six.
• If the hundred digit is 2, there are five.
• Et cetera, the answer is $6+5+\cdots+1=21$

I used C program as

include<stdio.h>

void main() {
int i, count = 0;
int r,n,a;
for(i=100; i<=600;i++) {
a = i;
n = 0;
while(a > 0) {
r = a%10;
n = n+r;
a/=10;
}
if(n == 6)
count++;
}
printf("%d", count);
}

and it prints 21

For every hundredth digit (n ≤ 6) there will be a (6-n) tenth digits So for 6(hundredth digit) - 1(possible case) for 5 - 2 . . for 1 - 6 So sigma n = n(n+1)/2 = 21

firstly fix hundredth digit as 1: 105,114,123,132,141,150 then fix hundredth digit a 2: 204,213,222,231,240 similarly 3: 303,312,321,330 4: 402,411,420 5: 501,510 6: 600

There is exactly twenty-one 3-digit numbers- 114, 141, 411, 105, 150, 510, 501, 123, 321, 231, 312, 213, 132, 600, 222, 240, 204, 402, 420, 330, 303.