My cryptogram hypothesis

Consider equations, an which A1,A2,A3,,An1,An{ A }_{ 1 },{ A }_{ 2 },{ A }_{ 3 },\dots { ,A }_{ n-1 },{ A }_{ n } is not necessarily distinct non-zero digits in decimal base and n,mn, m is natural numbers.

1. Prove for n5 n \ge 5 that equation below hasn't solution: A1+A1A2A2+A1A2A3A2A3++A1A2A3An1AnA2A3An1An=m\overline { { A }_{ 1 } } +\frac { \overline { { A }_{ 1 }{ A }_{ 2 } } }{ \overline { { A }_{ 2 } } } +\frac { \overline { { A }_{ 1 }{ A }_{ 2 }{ A }_{ 3 } } }{ \overline { { A }_{ 2 }{ A }_{ 3 } } } +\dots +\frac { \overline { { A }_{ 1 }{ A }_{ 2 }{ A }_{ 3 }\dots { A }_{ n-1 }{ A }_{ n } } }{ \overline { { A }_{ 2 }{ A }_{ 3 }\dots { A }_{ n-1 }{ A }_{ n } } } =m

2. Prove for n4 n \ge 4 that equation below hasn't solution: A1+A1A2A1A2+A1A2A3A2A1A2A3A2++A1A2An1AnAn1A2A1A2An1AnAn1A2=m\overline { { A }_{ 1 } } +\frac { \overline { { A }_{ 1 }{ A }_{ 2 }{ A }_{ 1 } } }{ \overline { { A }_{ 2 } } } +\frac { \overline { { { A }_{ 1 }{ A }_{ 2 }{ A }_{ 3 }{ A }_{ 2 }{ A }_{ 1 } } } }{ \overline { { { A }_{ 2 }{ A }_{ 3 }{ A }_{ 2 } } } } +\dots +\frac { \overline { { { A }_{ 1 }{ A }_{ 2 }\dots { A }_{ n-1 }{ A }_{ n }{ A }_{ n-1 }\dots { A }_{ 2 }{ A }_{ 1 } } } }{ \overline { { A }_{ 2 }\dots { A }_{ n-1 }{ A }_{ n }{ A }_{ n-1 }\dots { A }_{ 2 } } } =m

Note by Ilya Pavlyuchenko
5 months, 3 weeks ago

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Thank you, it is interesting

Emma Snow - 5 months, 1 week ago

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Isn't A(1)=1 and A(2)=2 and A(i)=0 for all i>2 a solution for equation 1?

Justin Travers - 5 months ago

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Whoops, just saw the requirement for non-zero digits.

Justin Travers - 5 months ago

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@Yajat Shamji, thank you for reporting the spam user. In the future, please tag my account so we can take care of the offending account as soon as possible.

Brilliant Mathematics Staff - 5 months ago

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@Ilya Pavlyuchenko, I don't know how to write cryptograms - I have got one for a problem but I can't write it. Can you give me any help on how to write cryptograms?

A Former Brilliant Member - 5 months, 2 weeks ago

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For working in LaTeX\LaTeX use plugin on browser Daum Equation Editor. You can download it free in Google Chrome Shop.

Ilya Pavlyuchenko - 5 months, 2 weeks ago

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Thanks for the alternative!

A Former Brilliant Member - 5 months, 1 week ago

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@A Former Brilliant Member No problem!

Ilya Pavlyuchenko - 5 months, 1 week ago

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@Brilliant Mathematics, I have noticed that in all the profiles of the spam users, the country they come from is Pakistan. Something relevant at all?

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No, I do not think so. But thanks for notifying us!

Brilliant Mathematics Staff - 5 months ago

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