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ASU 322 All Soviet Union 1981 greater gcd of each n,(n+1),...,(n+20) than 30030

Source:

July 23, 2019
number theorygreatest common divisorconsecutive

Problem Statement

Find nn such that each of the numbers n,(n+1),...,(n+20)n,(n+1),...,(n+20) has the common divider greater than one with the number 30030=2⋅3⋅5⋅7⋅11⋅1330030 = 2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13.
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