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32 = 25 What is the base in this number sentence?
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Is this using modular arithmetic?
\[32_k=25_m\] Note that k>3 and m>5. Convert to base 10: \[3k^1+2k^0=2m^1+5m^0\\ 3k+2=2m+5\\ 3k-2m=3\] Now we know that k has to be odd and m has to be a multiple of 3. After some trial and error, we find that the smallest integers k and m satisfying the above conditions are k=5, m=6.
nice one, @blockcolder ...
Thanks. :D
\[32_{2 n+3 }= 25_{3 (n+1)} \]where n, an integer, is >= 1. A solution using Mathematica Home Edition is attached.
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Sorry. Forgot to attached the pdf solution file.
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