Number Base Representation

This week, we learn about Number Base Representation.

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How would you convert (87654321)9 (87654321)_9 into base 3? Try and minimize the calculations.

Note by Calvin Lin
6 years, 1 month ago

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9=328=(2)3+(2)7=(2)3+(1)6=(2)3+(0)5=(1)3+(2)4=(1)3+(1)3=(1)3+(0)2=(0)3+(2)1=(0)3+(1)(87654321)9=(2)315+(2)314+(2)313+(1)312+(2)311+(0)310+(1)39+(2)38+(2)37+(1)36+(1)35+(0)34+(0)33+(2)32+(0)31+(1)30=(2221201211100201)3 \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

Siniša Bubonja - 6 years, 1 month ago

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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.

Calvin Lin Staff - 6 years, 1 month ago

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8765432187654321 becomes 807060504030201807060504030201 by only using 9=329=3^2. Furthermore 3n+a3n+a in base 99 is written as nana in base 33 and luckily nn is not greater than 22 because we have a number in base 99.

Applying this to 807060504030201807060504030201 = 8 07 06 05 04 03 0201 we get 22 21 20 12 11 10 0201 = 22212012111002012221201211100201

Ton de Moree - 6 years, 1 month ago

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

Snehal Shekatkar - 6 years, 1 month ago

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No problem :)

We look at the 77 for example. In base 99 it represents 7967*9^6. Using 9=329=3^2 we get 796=7(32)6=73127*9^6=7*(3^2)^6=7*3^{12}. With this we see that every power in base 99 is doubled when moving to base 33. Also, we see that this transformation only gives even powers of 33, hence all the zeroes on the places for odd powers of 33.

I hope this cleared things up :)

Ton de Moree - 6 years, 1 month ago

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@Ton de Moree Sorry but I still don't understand. 7967*9^6 is very big number. How come 7 is equal to 7*9^6 in base 9? I have studied different bases and all but this is very confusing to me.

Snehal Shekatkar - 6 years, 1 month ago

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@Snehal Shekatkar Ah, it's got to do with the position of the 77, not the 77 itself.

In base 1010 the number 321 means 3100+210+113*100 + 2*10 + 1*1 (multiples of powers of 1010.

In base 99 we use powers of 99, so 321 then means 381+29+113*81 + 2*9 + 1*1.

Now, in the number 8765432187654321 in base 99 means 897+796+695+594+493+392+291+1908*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 77 in (87654321)9(87654321)_9 stands for 7967*9^6.

Ton de Moree - 6 years, 1 month ago

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@Ton de Moree Thank you very much. I really did not see all these genius intricacies. :) But Still I am trying to understand your whole argument. It seems that you have taken too many short-cuts and for people like me it is very hard to understand your argument.

Snehal Shekatkar - 6 years, 1 month ago

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@Snehal Shekatkar It all makes sense in my head ;)

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

Ton de Moree - 6 years, 1 month ago

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@Ton de Moree No no.. I completely understand this.. For example.. as a chess player, I see 5-6 moves deep combination easily many times but my amateur friends often don't understand it without me explaining them all the elements. I am amateur in maths so I am just saying that you please describe your answer in little more detail. Thank you

Snehal Shekatkar - 6 years, 1 month ago

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

Philips Zephirum Lam - 6 years, 1 month ago

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

Justin Wong - 6 years, 1 month ago

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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(100.15)_{10} = (10201.01100..)_3 The .. in the end means there are more digits in the expansion.

Hariram K - 6 years, 1 month ago

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It's the same way, but it's much more complex / less applicable. Since 510.6310510.63_{10} implies 5(102)+1(101)+0(100)+6(101)+3(102)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(35)+2(32)+2(3)+1(31)+2(32)...=200220.12...32(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.

Michael Tong - 6 years, 1 month ago

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I think it has something to do with 9=32 9 = 3^{2}

Tan Li Xuan - 6 years, 1 month ago

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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=22212012111002013(87654321)_9 = |22|21|20|12|11|10|02|01|_3

8=2238 = |22|_3

7=2137 = |21|_3 and etc.

Tianbo Chen - 5 years, 9 months ago

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