scottishguy
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6÷2(1+2)= ?
Doing the rounds on loads of forums.
Do you get 1 or 9?
Doing the rounds on loads of forums.
Do you get 1 or 9?
BODMAS gives the order in which you do the calculations
Brackets - sort out what is inside the brackets first (obviously following the right order)
Orders - powers, square roots etc
Division, Multiplicationcan be done in any orderfrom left to right in the order of the sum)
Addition, Subtractioncan be done in any orderfrom left to right in the order of the sum
I don't think you can solve this without knowing if you multiply or divide what's produced either side of the bracket.
It should be 1.
(6/2)(1+2) = 9
6/2(1+2) = 1
You only multiply what is directly beside the bracket. If it's a bracket next to it then it's the whole sum. If not it's the adjacent number.
(6/2)/(1+2) = 1
Sorry Whitewash.. was trying to say that one side of the equation gives three. The bracket result gives three, but how do you decide if you multiply them or divide them. Is there a mathematical default?
but that is not the same sum, a number sat against a bracket such as 2(1+2) mean that it is multiply what is in the bracket by that number. in isolation this would give 6, however in our sum we have to do the preceeding divide 6/2x3 before multiplying the contents of the bracket with the two so we get 3x3 = 9

Err what about this then, is the site wrong:
http://www.mathsisfun.com/algebra/operations-order-calculator.html
This is the actual sum at the start.
oops might have given a bad example
I was implying the answer was 9 from the OPs question.
ziggy©;3631593 said:Based on BODMAS
Code:Brackets - 6/2(1+2) Multiplication - 6/2(3) Multiplication - 6/2*3 Division - 6/6 =1

DemiLion said:The answer's one.
I'd love to know how anyone is getting nine from six divided by six?