given a number X, find the lowest positive integer Y which can be devided by excactly X different integers (with no remainder).
For example, the number 20 can be divided by 6 different integers, namely 1, 2, 4, 5, 10 and 20. However, when the given number X is 6, Y should be 12, since 12 can be divided by 1, 2, 3, 4, 6 and 12 (also 6 in total), and there is no lower number with such many factors.
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This problem is much more interesting than the current test suite suggests, and could be made a lot more challenging with a larger number of test cases.
Agree with @Tim. It would be nice if at least 2, 9, 12, 15, 16, 18, 20, 21, 24, 25, 27, 28, 30, 32, 33, 35, 36, 39, 40, 42, 44, 45, 48, 49, 50, 51, 52, 54, 55, 56, 57, 60, 63, 64, 65, 66, 68, 69, 70, 72, 75, 76, 77, 78, 80, 81, 84, 85, 87, 88, 90, 91, 92, 93, 95, 96, 98 and 99 (just to name numbers that the current leader fails on below 100) would be added as test cases.