# converting bases? (math, numbers)

## converting bases? (math, numbers)

How would I convert 16 base 10 (normal) into base 5?

Would be you able to show me a solid way to convert various bases all the time?

(until, now, I have been using some self-logic-some-trial method in my head) and I've decided I have to find a reliable standard way to do this.

Thanks!

Would be you able to show me a solid way to convert various bases all the time?

(until, now, I have been using some self-logic-some-trial method in my head) and I've decided I have to find a reliable standard way to do this.

Thanks!

**karooomph**- Posts : 74

Join date : 2008-11-19

Age : 23

Location : ubseikastan

## Re: converting bases? (math, numbers)

Well, a number in a base, such 142 base 8, is basically just 1*8

So, when converting from one base to another, we are solving the equation: digit

This means when converting from some base to base 10 is easy because we can just plug in the appropriate digits, base, and powers into: digit

Converting from 10 to other bases is a bit trickier, since we're going from a number system we normally use and write in to one we don't. You can't just plug in the digits into digit

To solve an equation like this, we'll "fill" up each digit, starting from the ones place. First, take the remainder of 98/8 - the operator that returns the remainder of the division is known as the modulus operator, or simply mod, so 98 mod 8 = 2 (just take the decimal part of the division of 98/8, which is .25, and multiply it by 8). So, so far, we have 2 in the ones place.

Next we do 98 mod 8

Finally, we do 98 mod 8

So, now, in short, to convert 16 base 10 to base 5:

16 mod 5 = 1; digit in 1s place

(16 mod 5

31 is the final answer.

^{2}+4*8^{1}+2*8^{0}in base 10 (digit_{n}*base^{n}+digit_{n-1}*base^{n-1}+...+digit_{0}*base^{0}, where base^{0}would just equal 1).So, when converting from one base to another, we are solving the equation: digit

_{n}*base^{n}+digit_{n-1}*base^{n-1}+...+digit_{0}= digit_2_{n}*base_2^{n}+digit_2_{n-1}*base_2^{n-1}+...+digit_2_{0}This means when converting from some base to base 10 is easy because we can just plug in the appropriate digits, base, and powers into: digit

_{n}*base^{n}+digit_{n-1}*base^{n-1}+...+digit_{0}, and since we write numbers in a base-10 number system, this result is automatically in base 10 (so like in my example at the beginning, 142 base 8 = 1*8^{2}+4*8^{1}+2*8^{0}= 98 base 10).Converting from 10 to other bases is a bit trickier, since we're going from a number system we normally use and write in to one we don't. You can't just plug in the digits into digit

_{n}*base^{n}+digit_{n-1}*base^{n-1}+...+digit_{0}as we did before...but, for example, to convert 98 base 10 to base 8, we would have the equation: digit_{n}*base^{n}+digit_{n-1}*base^{n-1}+...+digit_{0}*base = 98.To solve an equation like this, we'll "fill" up each digit, starting from the ones place. First, take the remainder of 98/8 - the operator that returns the remainder of the division is known as the modulus operator, or simply mod, so 98 mod 8 = 2 (just take the decimal part of the division of 98/8, which is .25, and multiply it by 8). So, so far, we have 2 in the ones place.

Next we do 98 mod 8

^{2}, which equals 34, to see how much is in the eights place. Then, divide this result by 8, so 34/8=4.25. This means that 8 goes into 34 four whole times (the extra .25 is taken care of by the ones digit, 2, as 2/8=.25), so 4 is the the digit in the eights place, making our number so far: 42 base 8.Finally, we do 98 mod 8

^{3}, which actually equals 98 (this means we can stop the process of taking mods and dividing by 8^{power}here because 98 mod 8^{3}equalling 98 means that 8^{3}, which equals 512 base 10, is greater than 98). Then, we divide this result by 8^{2}to see how many times 8^{2}goes into it, or what the digit for the 8^{2}place would be (like how you 403 divided by 10^{2}equals 4.03, and 4 is the digit in the 10^{2}s place) - the result of this is 1.53125, so 1 is the digit in the 10^{2}s place, leaving us with the final answer of: 142 base 8 = 98 base 10.So, now, in short, to convert 16 base 10 to base 5:

16 mod 5 = 1; digit in 1s place

(16 mod 5

^{2}= 16; now do 16/5 to find how many times it can fit in the 5s place, 16/5=3.2, so 3 is the digit in the 5s place; we also knows this is the last digit we need to worry about because 16 mod 5^{2}= 1631 is the final answer.

**bfrsoccer**- Administrator
- Posts : 61

Join date : 2008-11-09

## Re: converting bases? (math, numbers)

Man. I'm glad I never have to go that far to answer a chemistry question. I don't know how you do it, bfr.... Nice use of sub and superscript.

**funion987**- Administrator
- Posts : 146

Join date : 2008-11-09

## speechless

to put in super and sub scripts, according to funion's info post you put these intriguing [sup] brackets. You did that at least 43 times in this detailed solution. That IS....dedication!

Pretty sure I won't have any more problems with bases from now on!!!

Thank you.

btw, gosh, I've been browsing through trying to answer any questions, but man, they're all like, "College Math, Riemann Theory, triflumetrapolimosiss! (you should get my point).

Tough.

But again, I will now always have this page for reference, thank you bfrsoccer!

**karooomph**- Posts : 74

Join date : 2008-11-19

Age : 23

Location : ubseikastan

## Re: converting bases? (math, numbers)

Unless I delete it....bwa ha ha ha.karooomph wrote:But again, I will now always have this page for reference, thank you bfrsoccer!

**funion987**- Administrator
- Posts : 146

Join date : 2008-11-09

## Re: converting bases? (math, numbers)

Good thinking.

**funion987**- Administrator
- Posts : 146

Join date : 2008-11-09

## Term Papers

I am trying to convert base 10 to base 16 and I can’t seem to wrap myself around how to do this. I understand how you did the base 5, but the answers that I am coming up with aren’t matching the book so I am also confused!

Term Papers

Term Papers

**orten999**- Posts : 19

Join date : 2010-04-01

## Re: converting bases? (math, numbers)

To solve an equation like this, we'll "fill" up each digit, starting from the ones place. First, take the remainder of 98/8 - the operator that returns the remainder of the division is known as the modulus operator, or simply mod, so 98 mod 8 = 2 (just take the decimal part of the division of 98/8, which is .25, and multiply it by . So, so far, we have 2 in the ones place.

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**LeeRain**- Posts : 12

Join date : 2010-09-10

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