Solution
Distinct Parts
Show the problem again
How many ways are there to write 20 as a sum of distinct positive integers, where the order of the summands does not matter? The one-term sum "20" counts. Give an integer.
Worked solution
**Answer: 64.** Count by the number of parts k. If the parts are a_1 > a_2 > ... > a_k, subtract k − 1, k − 2, ..., 0 from them respectively. This removes k(k − 1)/2 in total and leaves a partition of 20 − k(k − 1)/2 into exactly k positive parts (possibly repeated), and the map is reversible. - k = 1: 1 way. - k = 2: partitions of 19 into 2 parts: 9. - k = 3: partitions of 17 into 3 parts: 24. - k = 4: partitions of 14 into 4 parts: 23. - k = 5: partitions of 10 into 5 parts: 7. - k = 6 would need at least 1 + 2 + ... + 6 = 21 > 20. Total 1 + 9 + 24 + 23 + 7 = 64. By Euler's theorem this also equals the number of partitions of 20 into odd parts, which an independent enumeration confirms.
Source: Classic exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.