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Question: "Find all values of the digit m, if any, for which $m{\rm A} \mbox{ base fifteen} + m{\rm B} \mbox{ base fifteen} + m{\rm E} \mbox{ base fifteen} − \mbox{CD base fifteen} = 304 \mbox{ base } m.$"

To start off, I converted CD to base ten and got $193.$ I then converted $304$ base $m$ to base ten and got $3m^2+0m+4.$ I moved the $193$ over to the other side of the equation and got $m{\rm A} \mbox{ base fifteen} + m{\rm B} \mbox{ base fifteen} + m{\rm E} \mbox{ base fifteen} = 193m^2-189.$

Next, I added $m\rm A,$ $m\rm B,$ and $m\rm E$ to get the number "$(3m+2)5$" base fifteen, which I then converted to base ten and set equal to $193m^2-189.$ Then, I solved, but didn't get an integer as a result.

Could someone confirm if what I did was right / wrong?

Thanks (and sorry about the tags or lack thereof - bases wasn't one of the options and it didn't let me create a tag for it).

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You made a mistake when you "moved the 193 over". You should still have a $3m^2$ term. Also, it's best to convert everything to base ten as soon as possible. This yields: \begin{align*} (15m + 10) + (15m + 11) + (15m + 14) - (15 \cdot 12 + 13) &= 3m^2 + 4 \\ 45m - 158 &= 3m^2 + 4 \\ 0 &= 3m^2 - 45m + 162 \\ 0 &= m^2 - 15m + 54 \\ 0 &= (m - 6)(m - 9) \\ m &= 6,9 \end{align*}

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  • $\begingroup$ Sorry, that was my mistake when copying over to stack exchange. I still had am m^3 term, but I see I went wrong somewhere else along the line. Thanks for pointing that out! $\endgroup$ – Math Student Oct 27 '14 at 21:45

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