Do there exist 99 consecutive natural numbers such that the smallest one is divisible by 100, the next by 99, the third by 98, ..., and the last one by 2?
Source: Tournament of Towns - 2017 - O Level - 2
Yes. Exist.
Yes. (100!-n) for each n from 2 to 100 will do the trick. The set of all solutions can be found by replacing "100!" with "kM", where k is any natural number and M is the least common multiple of all numbers from 2 to 100.
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