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#1 |
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May 2004
New York City
5·7·112 Posts |
Define an "alternation number" as an integer whose (decimal)
representation contains exactly two different digit values, appearing in alternating place values (e.g. 1212 and 37373). Find an alternation number which factors into two (or more) alternation numbers each containing at least three digits (if possible). |
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#2 |
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Jan 2005
Transdniestr
503 Posts |
There's no solutions of the form: aba * cdc.
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#3 |
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6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2·7·19·37 Posts |
what about the forms: aba * aca , aba * cac , aba * bcb , aba * cbc ?
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#4 |
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Jan 2005
Transdniestr
50310 Posts |
None of them work.
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#5 |
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"Brian"
Jul 2007
The Netherlands
63058 Posts |
Tested all alternation numbers as far as 98,989,898,989,898,989
= 9*10^16 + 8*10^15 + ... + 9*10^0 None of the factorizations work. |
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#6 |
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"Brian"
Jul 2007
The Netherlands
CC516 Posts |
Oops. I have misread the puzzle.
My trivial effort above looked only at the prime factorizations of alternation numbers, but the problem does not state that the alternation factors have to be prime. Sorry. Keep trying. |
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