Tricimal

ryan_kidz

Junior Member
Joined
Sep 11, 2005
Messages
89
A tricimal is like a decimal, excet the digits represent fractions with powers of 3 instead of 10. FOr example, 16/27 = 1/3 + 2/9 +1/27 = 0.121 as a tricimal.
How is 77/81 expressed as a tricimal?

I got really confused with that problem. it looks not that complicated, but i still can't figure it out. :cry:

any brilliant people can help me?
 
Do you know how to convert bases?. Convert your numbers to from base 10 to base 3.

Let's do 77. Repeatedly divide by 3, keep track of remainders:

77/3=25 remainder 2

25/3=8, remainder 1

8/3=2, remainder 2


The base 3 representation of 77 is, from the top, equal to 2212.

81 is easy.

81/3=27, remainder 0

27/3=9, remainder 0

9/3=3, remainder 0

3/3=1, remainder 0

81=10,000
 
galactus said:
Do you know how to convert bases?. Convert your numbers to from base 10 to base 3.

Let's do 77. Repeatedly divide by 3, keep track of remainders:

77/3=25 remainder 2

25/3=8, remainder 1

8/3=2, remainder 2


The base 3 representation of 77 is, from the top, equal to 2212.

81 is easy.

81/3=27, remainder 0

27/3=9, remainder 0

9/3=3, remainder 0

3/3=1, remainder 0

81=10,000

I don't know how to convert bases :oops:

Thnx for the explaination.
 
Wait.. Can u explain to me how to convert bases?

I just want to know more about that.

thnx! :wink:
 
We really can't teach classes here. (We deal with specific questions on particular exercises.) Try doing a search, maybe on Google, for change-of-base lessons.

Eliz.
 
Hello, ryan_kidz!

How is \(\displaystyle \frac{77}{81}\) expressed as a tricimal?
We already have: .\(\displaystyle \L\frac{77}{3^4}\)

Express \(\displaystyle 77\) in base-3.. . \(\displaystyle \L77 \:= \:2\cdot3^3 + 2\cdot3^2 + 1\cdot3 + 2\cdot1\)

The fraction becomes: .\(\displaystyle \L\frac{2\cdot3^3\,+\,2\cdot3^2\,+\,1\cdot3\,+\,2\cdot1}{3^4}\;= \;\frac{2}{3^1} + \frac{2}{3^2} + \frac{1}{3^3} + \frac{2}{3^4}\)


Therefore: .\(\displaystyle \L\frac{77}{81} \;= \;0.2212_{_3}\)
 
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