# Combinatorics: How many 3-digit numbers have digits which have a sum of:

An 11th grade SAT questions asked students to find the number of 3-digit numbers that have a sum of 6. On this ubiquitous test that gains most students entrance to college, you only have a little over a minute to answer the question. By randomly finding solutions like 501 and 321 and 411 and 105 it would take 10 minutes and you would never solve this seemingly easy question.

When I posed this question to the Mathletes, from grades 2-7, they started out randomly as do most 11th graders, but after a few minutes of asking them if this is the best approach to find the solution, most of them made the abstract jump from a simple addition question to the field of combinatorics to which they were introduced last week. They started asking whether we could come up with a pattern or system of recording these solutions.

Here is the best practice solution:

105 204 303 402 501 600

114 213 312 411 510

123 222 321 420

132 231 330

141 240

150

Finding that there were 21 solutions was only the beginning. They saw patterns in each column, in each row, they noticed a triangular array of numbers, and that it was a sum of descending whole numbers.

We talked about how to define a three-digit number. Most had trouble articulating exactly what it was (most explanations spit back that it has three digits and when I wrote down “001” on the board, they then knew that the language of mathematics is very challenging; 001 is actually a one digit number).

We looked at the range of solutions from 100-999. Then, I asked the a simple question of how many three digit numbers are there? Most of them subtracted 100 from 999 to get 899. But this is a trap. There are actually 900 numbers; the easiest way to prove this is to look at how many whole numbers are there from 10-20. Most students will subtract 10 from 20 to get 10 solutions. However, when counting from 10 to 20 we see that there are 11 solutions. Then we looked at how many whole numbers there are from 10 to 12 and the answer is not 3. So, 999 less 100 is 899 but there are 900 solutions. Another proof is to look at how many whole numbers there are between 1 and 999; since there are 999 and if you subtract the 99 non-three-digit numbers you get 900.

Then I challenged them to explore all sums of three digit numbers from 1-27. In groups of 2 or 3, they started with a sum of 1 and recorded 1 solution (100). Then a sum of 2 with three solutions:

101 200

110

Then a sum of 3 with six solutions:

102 201 300

111 210

120

It was not long before they began to see a pattern in the number of solutions for each sum:

Sum of: Number of solutions: Sum of columns:

1 1 1

2 3 2 + 1

3 6 3 + 2 + 1

4 10 4 + 3 + 2 + 1

5 15 5 + 4 + 3 + 2 + 1

6 21 6 + 5 + 4 + 3 + 2 + 1

7 28 7 + 6 + 5 + 4 + 3 + 2 + 1

8 36 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1

9 45 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1

This is the sequence of triangular numbers. Once they found this pattern, they began to write down the solutions without solving the sum of three digit columns. I asked them to think about whether that pattern would continue indefinitely? When I suggested that the pattern breaks, they started to make conjectures (educated guesses) about where this would happen. They will find that the pattern breaks at a sum of 10. There are 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 solutions or 54 (there is no multiple of 100 in the solution set. Then a sum of 11 is 9 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 solutions or 61; then a sum of 12 has 66 solutions; a sum of 13 has 69 solutions and finally a sum of 14 has 70 solutions, the highest number. Then the number of solutions begins to decrease at a perfectly symmetrical pattern until you get to a sum of 27 with one solution (999).

With the older children we discussed whey 14 has the highest number of solutions. It is the “median” (middle number) between 1 and 27.

There are a host of discoveries here for each child.

Attachment | Size |
---|---|

Combinatorics_3-Digit_Sums.pdf | 52.64 KB |