The digits in the number, 42, add to six. There are exactly six 2-digit numbers with this property: 15, 24, 33, 42, 51, and 60.

How many 3-digit numbers exist for which the sum of the digits is six?

The answer is 21.

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Kenny Lau
- 6 years, 6 months ago

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 $6$ .

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Nycc one brian 👍

Aastha Shah
- 6 years, 6 months ago

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

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I used C program as

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

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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