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# 6÷2(1+2)=?

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posted on May, 1 2011 @ 07:46 PM
The answer is 1.

posted on May, 1 2011 @ 07:49 PM
reply to post by windwaker

My bad...Google says the answer is 9.

posted on May, 1 2011 @ 08:17 PM
I think the issue arises from how people visualize the problem.

For instance.

Someone, when reading" 6÷2(1+2)=?" can misinterpret it to be : "6÷(2(1+2)=?" which is exactly what happened to me.

However,

When you visualize like this:

, it makes more sense (to me at least).
edit on 1-5-2011 by AxlJones because: (no reason given)

posted on May, 1 2011 @ 08:22 PM
If you are getting "9", it is because you are not using distributive law. The answer is 1.

posted on May, 1 2011 @ 08:46 PM
reply to post by Cuervo

You are using the distributive property incorrectly. See my previous posts.

posted on May, 1 2011 @ 08:55 PM
reply to post by Cuervo

Your exactly right you have to implement the distributive law!

THE DISTRIBUTIVE LAW
For all real numbers a, b, and c:
a(b+c)=ab+ac
The distributive law offers two different orders of operation that always give the same result:

The expression a(b+c) specifies this order:
add b to c; then multiply a by this sum.
The expression ab+ac specifies this order:
multiply a and b; multiply a and c; then add these results.
One effective visual way to understand the statement of the distributive law is to use areas (see below).

So 6/2(1+2) Lets break it down! 6 divided by 2(1+2)
then gives you 6 divided by (2+4)

Now we are left with 6 dived by 6= 1

posted on May, 1 2011 @ 09:00 PM

Originally posted by ASeeker343
reply to post by Cuervo

You are using the distributive property incorrectly. See my previous posts.

Hmmm... I'm still standing by "1". I see what you are saying but I'm pretty sure that, resolved or not, you still need to use distributive law.

Then again, I always hated distributive law.

posted on May, 1 2011 @ 09:16 PM

Originally posted by lifeoflyman
reply to post by Cuervo

Your exactly right you have to implement the distributive law!

THE DISTRIBUTIVE LAW
For all real numbers a, b, and c:
a(b+c)=ab+ac
The distributive law offers two different orders of operation that always give the same result:

The expression a(b+c) specifies this order:
add b to c; then multiply a by this sum.
The expression ab+ac specifies this order:
multiply a and b; multiply a and c; then add these results.
One effective visual way to understand the statement of the distributive law is to use areas (see below).

So 6/2(1+2) Lets break it down! 6 divided by 2(1+2)
then gives you 6 divided by (2+4)

Now we are left with 6 dived by 6= 1

You got the distributive law correct there but you are implementing it wrong. a(b+c) >> 6÷2(1+2)

in this case its obvious what b and c are (b+c)=(1+2)

a however, is not soley 2. it is 6÷2

the distributive law does not state that you can just pull the two off away from the division sign and perform that operation first. 6÷2=3 which is equal to a in the algebraic example >> 3(1+2)=(3+6)=9

The issue that keeps arising is the parentheses. Because they are there does not mean the two is distributed before the division, it just specifies what is added.

posted on May, 1 2011 @ 09:19 PM

Originally posted by prolific
reply to post by Itop1

You wouldn't need Google if you knew that 2(3) is not distribution but its actually multiplication.

Heres an example:

edit on 1-5-2011 by prolific because: (no reason given)

This Exactly.

Good find!

posted on May, 1 2011 @ 09:37 PM

posted on May, 1 2011 @ 09:40 PM
reply to post by ASeeker343

6÷2(1+2)=

so according to the order of operations if you have 6*2÷3= You would do the multiplication first! According to your statement you say that since 2÷3 are together you can not break them apart there for that in of itself violates the orders of operation because you are multiplying only the 6 and the 2.

I think the problem is in how it is written. I would say that 6/2 and 6÷2 could be interpreted differently. as ÷ shows the action that is need to be done as / could be used for division or a fraction.

What if the problem was 6*2(1+2) what would be the first action you would take?

posted on May, 1 2011 @ 09:50 PM
The answer is clearly 1 no matter how you look at it. There is no difference in BEDMAS and PEMDAS, the order of similar level math operations, hence division and multiplication, doesn't matter. It also doesn't matter if you add or subtract first, as long as you perform both of those operations after the higher level operations of multiplication and division and exponents and brackets.

With the distributive property it's;

6÷2(1+2)

2(1)+2(2)=6

6÷6=1

With BEDMAS it's;

6÷2(1+2)

6÷2(3)

6÷6=1

It's no different with PEMDAS.

It is wrong to assume that an equation is solved from left to right like the reading of English words.

posted on May, 1 2011 @ 09:57 PM

Originally posted by snowspirit
6÷2(1+2)=?

Order of operations: BEDMAS

Brackets
Exponents
Division
Multiplication
Addition
Subtraction

1+2=3
2x3=6
6/6=1
edit on 1-5-2011 by snowspirit because: spelling

You didn't follow your own advice. It is 9.
2nd

posted on May, 1 2011 @ 10:03 PM

Originally posted by lifeoflyman
reply to post by ASeeker343

6÷2(1+2)=

so according to the order of operations if you have 6*2÷3= You would do the multiplication first! According to your statement you say that since 2÷3 are together you can not break them apart there for that in of itself violates the orders of operation because you are multiplying only the 6 and the 2.

I think the problem is in how it is written. I would say that 6/2 and 6÷2 could be interpreted differently. as ÷ shows the action that is need to be done as / could be used for division or a fraction.

What if the problem was 6*2(1+2) what would be the first action you would take?

According to the order of operations, ill use PEMDAS for simplicity here, Parentheses are first. This means the stuff inside the parentheses, not stuff adjacent to them. E- there are no exponents or powers to worry about. MD, multiplication and division are then handled from left to right. So for your example, I would add the 1 and 2 then multiply 6*2*3=36. Since the example you gave is all multiplication it doesnt really matter the order you do it in though. a(bc) = abc = (ab)c... its all the same. when division is implemented is when it gets tricky.

Your example could be done with the distributive property as well. 12(1+2) = 12+24 = 36
and since its all multiplication the order doesnt matter... 6*2(1+2) = 6*(2+4)=6*6=36
Since its all multiplication order doesnt matter.

posted on May, 1 2011 @ 10:05 PM
I know this is in General Chit Chat but seriously...

Here we are on a forum discussing intricate conspiracies, complex scientific conundrums, impossible social issues, and advanced metaphysics....and there is an argument over grade school math.

Or middle school math if you are in 2000.
Or high school math in 2010...

posted on May, 1 2011 @ 10:06 PM
Been corrected. It's 9.

It's misleading because the parentheses looks like it's bound to the 2.

Spread it out.

6 ÷ 2 (1+2)
6 ÷ 2 x (1+2)
6 ÷ 2 x (3)
6 ÷ 2 x 3 = ?????
edit on 1-5-2011 by Cataclysmo because: (no reason given)

posted on May, 1 2011 @ 10:07 PM
reply to post by Kody27

It is correct to solve left to right after following order of operations. In PEMDAS the M and D, multiplication and division are done at the same time before any addition or subtraction. Addition and subtraction are again done concurrently left to right. You dont go through and do all the addition first and then subtraction, same concept with multiplication and division.

[link]http://www.math.com/school/subject2/lessons/S2U1L2GL.html[link]

The correct answer is 9. You are multiplying out of order if you get 1.

posted on May, 1 2011 @ 10:08 PM
1.

Correction: I have to eat crow. The answer is 9.
edit on 1-5-2011 by alyoshablue because: (no reason given)

edit on 1-5-2011 by alyoshablue because: (no reason given)

posted on May, 1 2011 @ 10:10 PM

Originally posted by Cataclysmo
This is hilarious:

6÷2(1+2)=1

6 ÷ 2(1+2)
c ÷ b(a+b) (a+b) = d
c ÷ b(d)
c÷ bd

Brackets first, then if you remember how algebra works "bd" together means you multiply those two. And if I remember correctly because it's written as b(d) that means this has to be resolved first before you finish the rest. It's like exponents in that they are attached to each other. Would you do 6 ÷ 2^(3) with 6 divided by 2 first or complete finding out what 2 to the third power was?

False

Brackets first means you solve what is inside the brackes (1+2) the multiplication b(d) is an implied multiplication. Its the same as writing b*d

So... 6÷2(3) = 6÷2*3 not 6÷(2*3).

Once the argument inside the parentheses has been resolved they are irrelevant. You can go moving them around like that.

The answer is 9

posted on May, 1 2011 @ 10:14 PM

edit on 1-5-2011 by ASeeker343 because: (no reason given)

Remove post please
edit on 1-5-2011 by ASeeker343 because: (no reason given)

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