Yeah, all three of those are totally similar in ethical terms!

Walk away, into your pit of incorrect mathematical syntax.

Please don't get a job involving vector calculus, it will explode in your face.

I'm sick of trying to explain the rules around implied multiplication.
It should technically be 1, but 9 is also correct since the syntax of the question has been written non-determinately.

Read #307.

You're not applying the rules correctly which is why you're getting it wrong. You're making the same schoolboy error of classifying 2(1+2) as 2*(1+2) which is not correct shorthand if you're going to be using PEMDAS. The 2 is part of the parenthesis - it's a scaling factor.

Identically 6/(2+4) can have the denominator simplified by taking out a factor of 2 to 6/2(1+2) which by PEMDAS will lead you astray if you take it as now being a multiplication.

Walk away, into your pit of incorrect mathematical syntax.

Please don't get a job involving vector calculus, it will explode in your face.

That's the entire point. The 'arbitrary' rule isn't arbitrary. It's an explanation of how this syntax is written. Written division operator leading onto non-specified operator of 2(1+2) means that the 2(1+2) is grouped. This isn't up for discussion, this is real-life mathematical syntax. Had it been written legitimately as 6/2*(1+2) then you would be correct because NOW you would be performing it as it was written.

I'm sick of trying to explain the rules around implied multiplication.
It should technically be 1, but 9 is also correct since the syntax of the question has been written non-determinately.

Input error on user side. 8th time this has been pointed out now?

Be wary of relying of calculators, engage brain directly.

Read #307.

You're not applying the rules correctly which is why you're getting it wrong. You're making the same schoolboy error of classifying 2(1+2) as 2*(1+2) which is not correct shorthand if you're going to be using PEMDAS. The 2 is part of the parenthesis - it's a scaling factor.

Identically 6/(2+4) can have the denominator simplified by taking out a factor of 2 to 6/2(1+2) which by PEMDAS will lead you astray if you take it as now being a multiplication.

Walk away, into your pit of incorrect mathematical syntax.

Please don't get a job involving vector calculus, it will explode in your face.

Again, read #307. The syntax as written is what you're misinterpreting.

Inputting into a calculator is a direct result of the same mistake.

That's the entire point. The 'arbitrary' rule isn't arbitrary. It's an explanation of how this syntax is written. Written division operator leading onto non-specified operator of 2(1+2) means that the 2(1+2) is grouped. This isn't up for discussion, this is real-life mathematical syntax. Had it been written legitimately as 6/2*(1+2) then you would be correct because NOW you would be performing it as it was written.

I'm sick of trying to explain the rules around implied multiplication.
It should technically be 1, but 9 is also correct since the syntax of the question has been written non-determinately.

Not following general syntax. Check out #307.

Again, read #307. The syntax as written is what you're misinterpreting.

Inputting into a calculator is a direct result of the same mistake.

That's the entire point. The 'arbitrary' rule isn't arbitrary. It's an explanation of how this syntax is written. Written division operator leading onto non-specified operator of 2(1+2) means that the 2(1+2) is grouped. This isn't up for discussion, this is real-life mathematical syntax. Had it been written legitimately as 6/2*(1+2) then you would be correct because NOW you would be performing it as it was written.

I'm sick of trying to explain the rules around implied multiplication.
It should technically be 1, but 9 is also correct since the syntax of the question has been written non-determinately.