+1537 If $n$ is a positive integer, a $3$-partition of $n$ is a list of powers of $3$, arranged in order from greatest to least, whose sum is $n.$ For example, there are three $3$-partitions of $6,$ namely $3-3, 3-1-1-1$ and $1-1-1-1-1-1.$ How many different $3$-partitions of $8$ are there?
+1632 Let's list the 3-partitions of 8 out, since 8 is pretty small:
3-3-1-1
3-1-1-1-1-1
1-1-1-1-1-1-1-1 So in total there are THREE 3-partitions of 8.
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Now, if you are trying to find the # of 3-partitions of a number bigger than 8 (perhaps much larger), you can do casework (a strategy where you count the number of cases for each broad possibility) on the number of times the largest power of 3 appears, and then within that you can do another casework for the number of times the second largest power of 3 appears, so on so forth until you end up with 1s.
For example: # of 3-partitions of 28:
Case 1: One appearance of 3^3
27-1
Case 2.1: No 3^3, but two 3^2
9-9-3-3-3-1
9-9-3-3-1-1-1-1
9-9-3-1-1-1-1-1-1-1
9-9-1-1-1-1-1-1-1-1-1-1
Case 2.2: No 3^3, only one 3^2
9-3-3-3-3-3-3-1
Instead of listing out everything, note how a 3 can be expressed as 3 or 1-1-1, so you can either have six 3s, five 3s... 0 3s, total of 7 cases- a much better strategy and time-saver as to listing out all.
Case 3: No 3^3 or 3^2, only 3^1
3-3-3-3-3-3-3-3-3-1
You can have nine 3s, and a 1, eight 3s and four 1s... and counting the number of appearances of 3s, you can have nine appearances all the way to 0 appearances: 10 cases
Case 4: Only 1s.
Clearly, a 1-1-1-1-1-1... string of twenty-eight 1s.
In total, 1 + 4 + 7 + 10 + 1 = 23
Ooooo note how 28 - 5 = 23, and 8 - 5 = 3. Where you subtract 5 from the original number to get the number of 3-partitions... coincidence or not? >:) I leave you with this question parmen
