This week, we learn about Number Base Representation.

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How would you convert \( (87654321)_9 \) into base 3? Try and minimize the calculations.

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## Comments

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TopNewest\( \mathit{9=3^2}\\ \\ 8=(2)\cdot 3+(2)\\ 7=(2)\cdot 3+(1)\\ 6=(2)\cdot 3+(0)\\ 5=(1)\cdot 3+(2)\\ 4=(1)\cdot 3+(1)\\ 3=(1)\cdot 3+(0)\\ 2=(0)\cdot 3+(2)\\ 1=(0)\cdot 3+(1)\\ \\ (87654321)_9=(2)\cdot 3^{15}+(2)\cdot 3^{14}+(2)\cdot 3^{13}+(1)\cdot 3^{12}+(2)\cdot 3^{11}+(0)\cdot 3^{10}+(1)\cdot 3^9+(2)\cdot 3^8\\+(2)\cdot 3^7+(1)\cdot 3^6+(1)\cdot 3^5+(0)\cdot 3^4+(0)\cdot 3^3+(2)\cdot 3^2+(0)\cdot 3^1+(1)\cdot 3^0=\\(2221201211100201)_3 \)

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Good explanation of the step that you are using.

Note that this works only because 9 is a power of 3. In most other cases, it is advisable to convert to base 10, unless you are at ease with calculation in different bases.

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I knew most of this, but what I want to see is how to convert non-integers into different bases.

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It's the same way, but it's much more complex / less applicable. Since \(510.63_{10}\) implies \(5(10^2) + 1(10^1) + 0(10^0) + 6(10^{-1}) + 3(10^{-2})\), then in say base 3, this number would equal \(2(3^5) + 2(3^2) + 2(3) + 1(3^{-1}) + 2(3^{-2})... = 200220.12..._3\). You don't see non-integers in other bases often because it's just so awkward.

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If you want only an approximation upto a few digits, one way to do this by multiplying with appropriate power of base. Suppose you want the base 3 representation of 100.150 upto 5 trinary places. Then start by multipling 100.150 with 3^5. This is 24336.45. Now get 24336 in trinary. This will be 1020101100. So we can say \[(100.15)_{10} = (10201.01100..)_3\] The .. in the end means there are more digits in the expansion.

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Is 2221201211100201 in base 3 and 42 374 116 in base 10.

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\(87654321\) becomes \(807060504030201\) by only using \(9=3^2\). Furthermore \(3n+a\) in base \(9\) is written as \(na\) in base \(3\) and luckily \(n\) is not greater than \(2\) because we have a number in base \(9\).

Applying this to \(807060504030201\) = 8 07 06 05 04 03 0201 we get 22 21 20 12 11 10 0201 = \(2221201211100201\)

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First line itself I did not understand. Please explain. Thanks

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No problem :)

We look at the \(7\) for example. In base \(9\) it represents \(7*9^6\). Using \(9=3^2\) we get \(7*9^6=7*(3^2)^6=7*3^{12}\). With this we see that every power in base \(9\) is doubled when moving to base \(3\). Also, we see that this transformation only gives even powers of \(3\), hence all the zeroes on the places for odd powers of \(3\).

I hope this cleared things up :)

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In base \(10\) the number 321 means \(3*100 + 2*10 + 1*1\) (multiples of powers of \(10\).

In base \(9\) we use powers of \(9\), so 321 then means \(3*81 + 2*9 + 1*1\).

Now, in the number \(87654321\) in base \(9\) means \(8*9^7 + 7*9^6 + 6*9^5 + 5*9^4 + 4*9^3 + 3*9^2 + 2*9^1 + 1*9^0\).

So the \(7\) in \((87654321)_9\) stands for \(7*9^6\).

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It really isn't a problem for me to explain myself, so feel free to ask any questions you might have!

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I believe using this trick: https://brilliant.org/discussions/thread/investigation-faster-conversion-through-bases/?ref_id=54083

would also work out nicely. We would just have to work backwards. \((87654321)_9 = |22|21|20|12|11|10|02|01|_3\)

\(8 = |22|_3\)

\(7 = |21|_3\) and etc.

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I think it has something to do with \( 9 = 3^{2} \)

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