How many distinct 3-digit numbers have a digit sum of 11?

Answer 1

-8

Answer 2

First of all, I'll define the smallest 3 digit number as 100 (you cannot have 003 or 021, for example). If the first two digits are zero, then you cannot have a single digit sum of 11, anyway. I'll show with one leading zero at the end.

With hundreds digit as 9, then there are 3 possibilities: 902, 911, 920

With hundreds digit as 8, there are 4 possibilities: 803, 812, 821, 830

With hundreds digit as 7, there are 5 possibilities: 704, 713, 722, 731, 740

With 6, there are 6 possible: 605, 614, 623, 632, 641, 650

With 5, there are 7 possible: 506, 515, 524, 533, 542, 551, 560.

Continue on: 4 --> 8 possible; 3 --> 9 possible; 2 --> 10 possible; 1 --> 11 possible.

Add 3 + 4 + 5 + .... + 10 + 11 = 63.

Now if you want to take a leading zero, you cannot have 00n (with n a single digit), you cannot have 01n either. But you can have 029, and the rest are:

038, 047, 056, 065, 074, 083, 092. So 8 more possibilities = 71.

A 3-digits number, cannot start with a zero, so the smallest number can be 119 and the largest can be 920.

Since we have only 9 digits, and the numbers which start with 1 cannot have a zero digit, the possibilities are 1 to 9, that is 9 possibilities. Now we can think about numbers which start with 2, 3, 4, 5, 6, 7, 8, and 9.

11 =

2 + 9, 10 possibilities, from 0 to 9

3 + 8, 9 possibilities, from 0 to 8

4 + 7, 8 possibilities, from 0 to 7

5 + 6, 7 possibilities, from 0 to 6

6 + 5, 6 possibilities, from 0 to 5

7 + 4, 5 possibilities, from 0 to 4

8 + 3, 4 possibilities, from 0 to 3

9 + 2, 3 possibilities, from 0 to 2

So there are 61 possibilities (9 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3).