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I realized an unusual property with $6 _{10}$, $5 _{8}$ and $9 _{16}$.

What these have in common is when you multiply them to an even number, you get the same 1's digit.

Here it is for base 10.

$$6 \cdot 2 = 12$$ $$6 \cdot 4 = 24$$ $$6 \cdot 6 = 36$$ $$6 \cdot 8 = 48$$ $$6 \cdot 10 = 60$$

and so on. It works for base 16 and base 8. I would imagine it working for other powers too.

Here it is for base 16. (As I'm doing this I'm seeing the same values, but in a different base, interesting...)

$$9_{16} \cdot 2_{16} = 12_{16}$$ $$9_{16} \cdot 4_{16} = 24_{16}$$ $$9_{16} \cdot 6_{16} = 36_{16}$$ $$9_{16} \cdot 8_{16} = 48_{16}$$ $$9_{16} \cdot A_{16} = 5A_{16}$$ $$9_{16} \cdot C_{16} = 6C_{16}$$ $$9_{16} \cdot E_{16} = 7E_{16}$$

Why is that? Its something like with $( B / 2 ) + 1$ where $B$ is the base.

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Please retag and reformat my question as appropriate. –  Daniel A. White Sep 8 '11 at 12:38
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You're almost there. If n=2k is an even number less than base, what is the 1's digit of ((1/2 * base) + 1) * 2k? –  TonyK Sep 8 '11 at 12:44
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It is horribly uncouth to use an asterisk for ordinary multiplication in $\TeX$. I've changed it. –  Michael Hardy Sep 8 '11 at 12:48
    
Thanks. I'm not a TEX guy. –  Daniel A. White Sep 8 '11 at 12:49
    
I agree... but I'd be even more inclined to use the \cdot formulation; in my experience the \times is primarily for vector cross products whereas \cdot is for scalars. Not that \times is wrong or anything... :) –  AdamRedwine Sep 8 '11 at 13:37
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2 Answers

up vote 13 down vote accepted

$$((1/2)B+1)(2x)=xB+2x$$ where $B$ is the base and $2x$ is the even number.

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very interesting. thanks for enlightening me. –  Daniel A. White Sep 8 '11 at 18:57
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HINT $\ $ For every divisor $\rm\:d\:$ of $\rm\:B\:$ we have $\rm\:(B/d + 1)\:n\:d\ \equiv\ n\:d\pmod{B}\:.\:$ Rearranged, this boils down to $\rm\ n\:d\:(B/d)\equiv 0\pmod{B}\:,\:$ i.e. that $\rm\:B/d\:$ is a zero-divisor in $\rm\:\mathbb Z/B\: =\:$ integers modulo $\rm\:B\:.$

Generally if $\rm\:a\:$ is a zero divisor, i.e. if $\rm\:a\:c\: = 0,\ a,c\ne 0\:$ then $\rm\:a\:x = b\ \Rightarrow\ a\:(x+c) = b\:,\:$ so said linear equation does not have unique solutions.

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