Given the natural numbers bbb...b (n digits equal to b) and bbbb...b(m digits equal to b), show that the greatest common divisor of these numbers is bbb ... b (M digits equal to b, where M is the greatest common divisor of n and m).
gcd(bbb...b, bbb...b)
2
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number-theory
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0This is essentially a duplicate of the question linked in my answer. – 2010-11-16