There is a mistake in my previous post.

There are 50 groups of 10 consecutive integers ranging from 200 through 699. We can exclude the 25 groups that have the forms a0b, a1b, a7b, a8b, and a9b. That leaves 25 groups. We can also exclude the five groups that have two multiples of 9 because those multiples end in 0 and 9. Those groups start at 270, 360, 450, 540, and 630. But I missed a double counting problem: we already excluded the 270 group. So we subtract 4 from 25, leaving 21 groups.

The next exclusion has a double counting problem that I did not miss. We need to exclude the two groups that follow a group containing two multiples of 9 because the multiples of 9 in those groups have a final digit of 8 or 7. That is 10 groups starting with 280, 290, 370, 380, 460, 470, 550, 560, 640, and 650. However, we previously excluded the groups starting with 280, 290, 370, 380, and 470. So we are only excluding 5, reducing 21 to 16.

Finally, w need to exclude the group that precedes a group containing two multiples of 9 because the multiples of 9 in those groups have a final digit of 1. Obviously, there are five such groups, namely those starting with 260, 350, 440, 530, and 620. There is no double counting issue here. So we are left with 11 groups, namely

220-229, 230-239, 240-249, 250-259, 320-329, 330-339, 340-349, 420-429, 430-439, 520-529, 660-669.

Sorry about that.

The 11 sums that are divisible by 9 are

225, 234, 243, 252, 324, 333, 342, 423, 432, 522, and 666.

I at first tried to find the different ways to put together pairs of numbers from the allowed set that sum to one of those eleven numbers by hand. It was tedious, and I kept making mistakes. There is sort of a pattern, but it was not obvious to me until I resorted to computer, which is how I found my counting mistake in the first place.

Here are the results.

possibilities for

225 are two, namely 112 + 113, 113 + 112

234 are six, namely 111+123, 112 + 122, 113 + 121, 121 + 113, 122 + 112, 123 + 111

243 are six, namely 111 + 132, 112 + 131, 121 + 122, 122 + 121, 131 + 121, 132 + 111

252 are two, namely 131 + 121, 121 + 131

324 are six, namely 111 + 213, 112 + 212, 113 + 211, 211 + 113, 212 + 112, 213 + 111

333 are eight, namely 111 + 222, 112 + 221, 121+ 212, 122 + 211, 211 + 122, 212 + 121, 221 + 112, 222 + 111

342 are six, namely 111 + 231, 121 + 221, 131 + 211, 211 + 131, 221 + 121, 231 + 111

423 are six, namely 111 + 312, 112 + 311, 211 + 212, 212 + 211, 311 + 112, 312 + 111

432 are six, namely 111 + 321, 121 + 311, 211 + 221, 221 + 211, 311 + 121, 321 + 111

522 are two, namely 311 + 211 , 211 + 311

666 is one, namely 333 + 333

Thus, unless I have made another error, there are exactly 51 six-digit numbers that contain only the digits 1, 2, and 3.

No way a kid of 11 is supposed to solve this unless this is for a programming class.