im doing a different homework now too and it’s using 224/3 in base 5

the answer is 23

i just dont know what the 3rd place value is

2 = ?
2 = 5
4 = 1

i know you just add up the values to find out what it is so its 4+10+? i thought the 3rd value was 25 but thats wrong

no it IS 25 because the highest value in base 5 is “4” so once you have 5 “5’s” you have to carry over just like when you have 10 of something in our system you carry over to the next place value so WTF? why doesnt this work out, what am i missing?

im doing a different homework now too and it’s using 224/3 in base 5

the answer is 23

i just dont know what the 3rd place value is

2 = ?
2 = 5
4 = 1

i know you just add up the values to find out what it is so its 4+10+? i thought the 3rd value was 25 but thats wrong

no it IS 25 because the highest value in base 5 is “4” so once you have 5 “5’s” you have to carry over just like when you have 10 of something in our system you carry over to the next place value so WTF? why doesnt this work out, what am i missing?[/quote]

When you say 224/3 = 23, are both sides of the expression in base 5 or is the left in base 5 and the right in base 10?

Doing long division in base 5 works fine too. 10/4 gives 1 remainder 1. Bring down the 3. 13/4 gives 2. Bring down the 4. 4/4 is 1. Thus your answer is 121.

If it is really 1032, you would have 120 remainder 2. If the teacher wants decimals instead of remainders, you would add a decimal and a zero to 1032, making it 1032.0, same thing. Your mistake was that you assumed 4 went into 20 fives times, and then converted it back to 10. Remember that 2.0 is still base 5, and when dividing it you treat is as 20, or 10 in base 10. Therefore, 20/4 is 2 remainder 2. Obviously the 2 will then continue forever.

im doing a different homework now too and it’s using 224/3 in base 5

the answer is 23

i just dont know what the 3rd place value is

2 = ?
2 = 5
4 = 1

i know you just add up the values to find out what it is so its 4+10+? i thought the 3rd value was 25 but thats wrong

no it IS 25 because the highest value in base 5 is “4” so once you have 5 “5’s” you have to carry over just like when you have 10 of something in our system you carry over to the next place value so WTF? why doesnt this work out, what am i missing?[/quote]

If you want to convert it to base 10 first, you just do a sum of powers.

2(5)^2+2(5)^1+4(5)^0=64

64/3 is 21 1/3. You then would have to convert that back to base 5.

You are better off just doing long division and remembering you are in base 5.

: 41 R1
: 3|224
: -22
: 04
: - 3
: 1

EDIT: My long division didn’t come out right. Answer is 41 R1.

EDIT EDIT: The colons should help get my point across.

EDIT EDIT EDIT: Maybe not. Hopefully you get the point. I can’t get the long division to indent so that it is readable.

im doing a different homework now too and it’s using 224/3 in base 5

the answer is 23

i just dont know what the 3rd place value is

2 = ?
2 = 5
4 = 1

i know you just add up the values to find out what it is so its 4+10+? i thought the 3rd value was 25 but thats wrong

no it IS 25 because the highest value in base 5 is “4” so once you have 5 “5’s” you have to carry over just like when you have 10 of something in our system you carry over to the next place value so WTF? why doesnt this work out, what am i missing?

When you say 224/3 = 23, are both sides of the expression in base 5 or is the left in base 5 and the right in base 10?[/quote]

they are both in base 5

and the answer turns out to be 41 (5) R1 (5)

i was looking at the wrong answer key…fuck my life.

nvm, im just a retard who was using the wrong chart somehow i was looking at 22 as 11 when in fact it is 12, it makes perfect sense now and took 2 seconds to do.

[quote]malonetd wrote:
If you have Excel, it’s pretty easy to make yourself a base N converter.[/quote]

nah i dont have it, but i think we’re supposed to learn how to use it in this course.

its a cool math course (did i just say that?) cause its not about learning new equations its really just about logic. its like a college level math course from one of those books youd pick up for a 10 hour flight.

out of curiosity though i wanna check one of my problems with you guys.

“You have a glass of water and a glass of root beer. ou transfer a tablespoon of root beer into the glass of water then a tablespoon of that mixture back into the root beer. Is there more root beer in the water or more water in the root beer or equal amounts in each? Explain your answer”

my answer was they are equal

explanation:

let 100 represent the the root beer and 100 represent the water

You are correct that they are equal, but I’m not following your logic.

Assume 100 units of root beer and 100 units of water. Let a tablespoon be one unit.

1 unit of root beer is placed in the water.
The water/rb mixture now has 101 units. A tablespoon of this mixture will be representative of the entire thing. It is therefore 100/101=~.99 units water and 1/101=~.01 units root beer. This leaves you with 99.01 units water and .99 units root beer in this glass.

Add this back into the root beer and the root beer now has ~99.01 units rootbeer and .99 units water.

It’s actually a very simple problem when you take a step back and consider it. If there is equal amount rootbeer and water to begin with, then no matter how you divy it up there will be equal amounts root beer and water when you finish. As long as the total volume of liquid is the same in both glasses, then rootbeer in glass A will always equal water in glass B, and vice versa.

[quote]tedro wrote:
You are correct that they are equal, but I’m not following your logic.

Assume 100 units of root beer and 100 units of water. Let a tablespoon be one unit.

1 unit of root beer is placed in the water.
The water/rb mixture now has 101 units. A tablespoon of this mixture will be representative of the entire thing. It is therefore 100/101=~.99 units water and 1/101=~.01 units root beer. This leaves you with 99.01 units water and .99 units root beer in this glass.

Add this back into the root beer and the root beer now has ~99.01 units rootbeer and .99 units water.

It’s actually a very simple problem when you take a step back and consider it. If there is equal amount rootbeer and water to begin with, then no matter how you divy it up there will be equal amounts root beer and water when you finish. As long as the total volume of liquid is the same in both glasses, then rootbeer in glass A will always equal water in glass B, and vice versa.

[/quote]

what youre saying looks like what im saying just way more complicated.