6 ÷ 2(1+2)

Completely off the Topic of Tea

6 ÷ 2(1+2)

0
1
2%
1
20
49%
2
0
No votes
3
0
No votes
4
0
No votes
5
0
No votes
6
0
No votes
7
0
No votes
8
0
No votes
9
20
49%
10
0
No votes
 
Total votes: 41

6 ÷ 2(1+2)

Postby auhckw » Sep 23rd, '11, 05:18

Bored and found this math question... Lets get back to school... what is the answer for...

6 ÷ 2(1+2)

?

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Re: 6 ÷ 2(1+2)

Postby auhckw » Sep 23rd, '11, 07:04


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Re: 6 ÷ 2(1+2)

Postby Fabien » Sep 23rd, '11, 15:57

The answer shown in ImageShack is wrong :)
In this equation, priority is given to the operation 2(1+2) before 6÷2.
So, the right answer should be 1.

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Re: 6 ÷ 2(1+2)

Postby AdamMY » Sep 23rd, '11, 16:00

Fabien wrote:The answer shown in ImageShack is wrong :)
In this equation, priority is given to the operation 2(1+2) before 6÷2.
So, the right answer should be 1.



Sorry but recall the order of operations:

1. Parenthesis
2. Exponents
3. Multiplication/Division which ever occurs first from left to right
4. Addition/Subtraction which ever occurs first from left to right.

so you start by adding (1+2)

and it turns into 6÷2(3).

But then as all that is left is multiplication or division, we look from left to right and 6÷2 goes first turning it into 3 times 3, or 9.

I should say rarely would someone in real life write a problem like that, and problems of that nature are supposed to be intentionally misleading.

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Re: 6 ÷ 2(1+2)

Postby Drax » Sep 23rd, '11, 16:30

AdamMY wrote:I should say rarely would someone in real life write a problem like that, and problems of that nature are supposed to be intentionally misleading.


Exactly.

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Re: 6 ÷ 2(1+2)

Postby bearsbearsbears » Sep 23rd, '11, 20:48

You can also think of it fractionally as (6(1+2))/2

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Re: 6 ÷ 2(1+2)

Postby Running for Tea » Sep 23rd, '11, 21:21

I couldn't tell you the last time I saw the ÷ sign in an equation . My initial instinct was 1, but caught myself... tricky tricky :?

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Re: 6 ÷ 2(1+2)

Postby fdrx » Sep 23rd, '11, 21:54

without those "spaces" around the "÷" it's less tricky :oops:

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Re: 6 ÷ 2(1+2)

Postby SlientSipper » Sep 24th, '11, 01:24

As I recall, the order of Operations went.

Dear
Ben
Please
Excuse
My
Dear
Aunt
Sally.

Divide
Bracket
Parenthesis
Exponents
Mutiplication.
Duh.... :shock:
Ahhhhhh :x
Sheesh... I'm stupid. :cry:

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Re: 6 ÷ 2(1+2)

Postby Xell » Sep 26th, '11, 08:36

Interesting, i didn't have second thoughts about answer 9 :D

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Re: 6 ÷ 2(1+2)

Postby CasualConnoisseur » Sep 27th, '11, 18:55

I would add that the division symbol confused me initially. Writing it out as fractions clears the ambiguity up; I guess that's why you stop using it once you hit Algebra 1 in high school. fdrx is right in that the spaces around the division symbol don't help clarity at all.

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Re: 6 ÷ 2(1+2)

Postby p_funk » Oct 18th, '11, 12:08

I can't believe that i got it right, much less remembered how to do it, lol.

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Re: 6 ÷ 2(1+2)

Postby bagua7 » Oct 18th, '11, 21:30

Let's keep the ball rolling:

The prize money of $180 from a competition is to be divided between three children.

Adam won the first prize, Beth won the second prize and Chris won the third prize.

It is decided that the division will be in the ratio of:

1st : 2nd : 3rd = 5 : 3 :1

How much does each child receive?



Answer: http://imageshack.us/photo/my-images/16/answert.jpg/

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Re: 6 ÷ 2(1+2)

Postby AdamMY » Oct 18th, '11, 22:26

If you are going to discuss math, at least ask interesting questions. Such as stuff relating to what I have been working on, but accessible to lay people who want to think logically.

Assuming between two people they either know each other or they do not. What is the smallest group of people you must assemble such that you either have 3 of them which all know each other, or 3 of them that do not know each other?

Similarly, what if we replaced that with either 4 people that all know each other, or 3 people that do not know each other?


(Hint before considering the second question tackle the first).

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Re: 6 ÷ 2(1+2)

Postby entropyembrace » Oct 19th, '11, 01:28

AdamMY wrote:If you are going to discuss math, at least ask interesting questions. Such as stuff relating to what I have been working on, but accessible to lay people who want to think logically.

Assuming between two people they either know each other or they do not. What is the smallest group of people you must assemble such that you either have 3 of them which all know each other, or 3 of them that do not know each other?


I don´t understand how this is possible to answer....if you have any number of people all of them could know each other or none of them could know each other or any number of pairs of people within the group could know each other.

At first I thought the answer was 3...because it´s certainly possible to have a group of 3 that knows each other or a group of 3 that does not know each other but it´s also possible to have a group of 3 in which one person knows both others but they do not know each other or a group of 3 in which two people know each other but neither knows the 3rd person.

If I make the group size 4 then if none of them know each other there are 4 people that do not know each other not 3 people that do not know each other. Any group size greater than 3 has this problem.

A group size less than 3 can´t satisfy the conditions in any case at all since there aren´t 3 people to know or not know each other.

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