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Previously divisible (
Posted on 20200924
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2020 positive integers are written in one line. Each of them starting with the third is divisible by previous and by the sum of two previous numbers. What is the smallest value the last number can take?
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Submitted by
Danish Ahmed Khan
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Solution?
 Comment 1 of 2
If I try to minimize each term in the series, one term at a time, I think the first 6 terms are 1,1,2,6,24,120, which looks like (n1)!
So I guess the solution is 2019!
Posted by
Steve Herman
on 20200924 16:05:15
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