You need 10 as a sum of multiples (0 to 2) of powers of 3. Since 10 is 9+1 and 9 is 3², in base 3 10 is 101, 1•3² + 0•3 + 1.
2006-12-17 17:00:31
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answer #1
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answered by Philo 7
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It has been awhile since I've done this, but basically I'll explain our base 10 number system and hopefully that will help. I hope I explain this right.
our numbers columns work on a base 10 system, meaning that
10^0 (the carret represents exponent) = 1 or the ones column in any number system, the 0 power is the one's column.
10^1 = 10
10^2 = 10 x 10 or 100
notice how the decimal places shift over one....now I'll try to make that into a base 3 system....
3^0 = 1 (like I said, any number to the 0 power is going to give you the one's column) This is your first decimal place
3^1 = 3 - this is your second decimal place
3^2 = 9 - this is your third decimal place
so now, just as we "group" things in terms of 10 in a base 10 system, you group them to what I said above.
1 = 1
2 = 2
3 = 10 (the 1 is 3^1 , there are no "ones" )
4 = 11 (the left "1" is 3^1 or 3, the right "1" is the ones place (3^0)
5 = 12
6 = 20
7 = 21
8= 22
9 = 100 (the one represents 3^2, there are no other decimal places)
10 = 101 (the one represents 3^2, no 3^1, and one 3^0 decimal place
Hope that helps. I think I did it right. Like I said, it's been awhile.
2006-12-17 17:15:00
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answer #2
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answered by coridroz 3
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Converting from base 10 to any other base number is very simple...
First you have to understand what each place in a number stands for. For example in the number 1234,
the digit "4" = 4 * 10^0
the digit "3" = 3 * 10^1
the digit "2" = 2 * 10^2
and the digit "1" = 1 * 10^3
let's say that we want to convert that into base 7... you divide the base 10 number by 7
1234/7 = 176 with a remainder of 2... that means that 2 is left over after the multiples of 7 are removed, so the 2 goes into the ^0 column
so we have one digit in the base 7 number, and we know that it equals:
0000002 base 7 + 176/7 base 10
176/7 = 25 with a remainder of 1... and the one is what's left over after the multiples of 7 are removed. Now we have two digits.
00000012 base 7 + 25/7 base 10
25/7 = 3 with a remainder of 4.... so....
0000412 base 7 + 3/7 base 10
3/7 = 0 with a remainder of 3... and this is our final digit in base 7
00003412 base 7 and we don't need the extra zeros, so our answer is
1234 base 10 = 3412 base 7
It's always a good idea to check our work... so...
2 = 2 * 7^0 = 2 * 1 = 2
1 = 1 * 7^1 = 1 * 7 = 7
4 = 4 * 7^2 = 4 * 49 = 196
3 = 3 * 7^3 = 3 * 343 = 1029... and we add those up
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= 1234 wow!!!
You can convert from base 7 back into base 10 by dividing just like we did converting from base 10... but math functions in different bases are a bit more complicated than what you are used to in base 10... so we'll save that for another time.
2006-12-17 17:49:16
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answer #3
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answered by sagacity_ron 2
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This Site Might Help You.
RE:
How do I represent the decimal number 10 in base 3?
I spent most of the day figuring out how to convert the decimal number to binary (base 2), to hexadecimal (base 16) and back to decimal (base 10). As if that was not a big enough pain, now I have to represent the decimal number 10 in base 3. The book does not cover base 3, neither does the web site...
2015-08-18 17:14:20
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answer #4
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answered by Gavra 1
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10/3 = 3 and the balance is 1
3/3 = 1 balance is 0
1/3 = 0 balance is 1
so the decimal no 10 in base 3 can we get from reading the balance from last to the first => 101
(this only work when you want to convert only decimal no to any base)
2006-12-17 20:11:34
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answer #5
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answered by Leh 2
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The main thing to realize is the place value of each digit in the base you are using. In base 10, from right to left, you have the ones place, tens place, hundreds place, etc. In base 3, the place value names would be ones, threes, nines, twenty-sevens, etc ... The decimal number 10 would require a 1 in the nines place, a zero in the threes, and a 1 in the ones place - 101. 25 would be a 2 in the nines, 2 in the threes, and 1 in the ones - 221. Hope that helps!
2006-12-17 17:17:49
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answer #6
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answered by JerryTcher 2
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For the best answers, search on this site https://shorturl.im/avyCj
Numbering systems work really easily. Take the system we use, base-10: 15563 What number is that? How do you know it's value just by looking at it? Can we break the number down? 1 * 10^4 + 5 * 10^3 + 5 * 10^2 + 6 * 10^1 + 3 * 10^0 All numbering systems work the same way. In any base-k system, a number can represented as: a(1) * k^0 + a(2) * k^1 + a(3) * k^3 + .... + a(n) * k^n where a(t) is an integer, and 0 = a(t) = k - 1 Now, a hexadecimal system is a base 16 system, with the following values (hexadecimal on left, decimal on right): 0 = 0 1 = 1 2 = 2 3 = 3 4 = 4 5 = 5 6 = 6 7 = 7 8 = 8 9 = 9 A = 10 B = 11 C = 12 D = 13 E = 14 F = 15 Now, your decimal number is 757.25. The first thing you need to do is ask yourself, "What's the first power of 16 that is greater than 757.25?" The answer would be 16^3 (4096 in decimal). In hexadecimal, this would be represented as 1000. So, we need to shoot just below that power, so we're looking at 16^2 (256 in decimal, 100 in hexadecimal). Now, how many times can 757.25 be divided by 256? 2 times. So, our first digit is 2. Now, 757.25 - 2 * 16^2 = 757.25 - 512 = 245.25. We go one step below and ask how many times 245.25 is divided by 16? 15. So, our second digit is F. 245.25 - 16 * 15 = 245.25 - 240 = 5.25 How many times can 5.25 be divided by 16^0 (1)? 5. Third digit is 5. Remainder is .25 Now, .25 = a * (1/16) .25 * 16 = a. 4 = a, with no remainder So, our number is: 2F5.40 Binary is easier still 0 = 0 1 = 1 2^k = 757.25 9 < k < 10 757.25 - 2^9 = 757.25 - 512 = 245.25 First digit is 1 Can 2^8 go into 245.25 at least once? no. 2nd digit is 0 Can 2^7 go into 245.25 at least once? yes. 3rd digit is 1. 245.25 - 128 = 117.25 Can 2^6 go into 117.25 at least once? yes. 4th digit is 1. 117.25 - 64 = 53.25 Can 2^5 go into 53.25 at least once? yes. 5th digit is 1. 53.25 - 32 = 21.25 Can 2^4 go into 21.25 at least once? yes. 6th digit is 1. 21.25 - 16 = 5.25 Can 2^3 go into 5.25 at least once? no. 7th digit is 0. Can 2^2 go into 5.25 at least once? yes. 8th digit is 1. 5.25 - 4 = 1.25 Can 2^1 go into 1.25 at least once? no. 9th digit is 0 Can 2^0 go into 1.25 at least once? yes. 10th digit is 1. 1.25 - 1 = .25 Can 2^-1 go into .25 at least once? no. 11th digit is 0. Can 2^-2 go into .25 at least once? yes. 12th digit is 1. .25 - 2^-2 = 0. No more remainders. So let's look at our number in binary: 1011110101.01
2016-04-04 02:49:53
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answer #7
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answered by Terri 4
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101 - 1, 2, 10, 11, 12, 20, 21, 22, 100, 101
2006-12-17 16:57:03
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answer #8
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answered by Anonymous
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