A Great Greatest Common Divisor Problem

How many digits does gcd ( 1111...1111 $240$ times , 1111...1111 $150$ times ) \text{gcd} (\underbrace{1111...1111}_{\text{\$240\$ times}},\underbrace{1111...1111}_{\text{\$150\$ times}}) have?

30 120 60 90

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1 solution

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Jul 2, 2018

gcd ( 1111...1111 $240$ times , 1111...1111 $150$ times ) \text{gcd} (\underbrace{1111...1111}_{\text{\$240\$ times}},\underbrace{1111...1111}_{\text{\$150\$ times}})

= gcd ( 1111...1111 $240$ times 1111...1111 $150$ times , 1111...1111 $150$ times ) =\text{gcd} (\underbrace{1111...1111}_{\text{\$240\$ times}}-\underbrace{1111...1111}_{\text{\$150\$ times}},\underbrace{1111...1111}_{\text{\$150\$ times}})

= gcd ( 1111....1111 $90$ times 0000....0000 $150$ times , 1111...1111 $150$ times ) =\text{gcd}(\underbrace{1111....1111}_{\text{\$90\$ times}}\underbrace{0000....0000}_{\text{\$150\$ times}},\underbrace{1111...1111}_{\text{\$150\$ times}})

= gcd ( 1111....1111 $90$ times , 1111...1111 $150$ times ) =\text{gcd}(\underbrace{1111....1111}_{\text{\$90\$ times}},\underbrace{1111...1111}_{\text{\$150\$ times}})

= gcd ( 1111....1111 $90$ times , 1111...1111 $60$ times ) =\text{gcd}(\underbrace{1111....1111}_{\text{\$90\$ times}},\underbrace{1111...1111}_{\text{\$60\$ times}})

= gcd ( 1111....1111 $30$ times , 1111...1111 $60$ times ) =\text{gcd}(\underbrace{1111....1111}_{\text{\$30\$ times}},\underbrace{1111...1111}_{\text{\$60\$ times}})

= gcd ( 1111....1111 $30$ times , 1111...1111 $30$ times ) =\text{gcd}(\underbrace{1111....1111}_{\text{\$30\$ times}},\underbrace{1111...1111}_{\text{\$30\$ times}})

= 1111...1111 $30$ times =\underbrace{1111...1111}_{\text{\$30\$ times}}

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