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Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?

I thought it would be 12 this as per pemdas rule:

$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$

Wanted to confirm the right answer from you guys. Thanks for your help.

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    $\begingroup$ linear-algebra? modular-arithmetic? Why did someone edit those tags in?? $\endgroup$ – mrf May 29 '13 at 9:03
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    $\begingroup$ @Matt: I don’t know, unless there are folks who think that being confused deserves a downvote. I certainly don’t, and have given it a compensatory upvote. $\endgroup$ – Brian M. Scott May 29 '13 at 9:04
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    $\begingroup$ Dear downvoters: I'm not sure that you should downvote simply because the OP was wrong. I don't quite see what's so bad about this question. It contains a simple, clear question with a straightforward answer, to which the OP offered their attempt in good faith. It also seems no more specific than any other question, and AFAIK is in the right place. $\endgroup$ – not all wrong May 29 '13 at 9:05
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    $\begingroup$ @Sharkos "Dear" downvoters? : ) $\endgroup$ – Rudy the Reindeer May 29 '13 at 9:06
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    $\begingroup$ Hail @Matt: It's a form of address, a salutation. If politeness is banned from M.SE, my good man, do direct me to the relevant meta page. Yours sincerely, Carl Turner, Esq. PS: Biting sarcasm actually might be, in which case I'm royally screwed. $\endgroup$ – not all wrong May 29 '13 at 9:11
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Your answer of $12$ is numerically correct, but you’ve applied the precedence rules incorrectly. In the absence of parentheses multiplication is done before addition, so

$$1+1+1+1+1+1+1+1+1+1+1+1\cdot 0+1$$

is to be evaluated as

$$1+1+1+1+1+1+1+1+1+1+1+(1\cdot 0)+1\;,$$

not as

$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot(0+1)\;.$$

Since $1\cdot0=0$, this simplifies to

$$1+1+1+1+1+1+1+1+1+1+1+0+1=12\;.$$

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Multiplication takes priority over addition, so this

$1+1+1+1+1+1+1+1+1+1+1+1\times0+1$

becomes:

$1+1+1+1+1+1+1+1+1+1+1+0+1$

Now add everything which gives you 12.

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If you type $1+1+1+1+1+1+1+1+1+1+1+1⋅0+1$ into most pocket calculators you will get the answer $1$. This is because the calculator applies the next operation to the result of the previous one, effectively computing$$(1+1+1+1+1+1+1+1+1+1+1+1)⋅0+1=0+1=1$$

When using a calculator you have to respect the logic of the machine. When writing mathematics you need to be aware of the standard conventions for brackets, which give the answer $12$ in this case. I suspect that the confusion arises because the two are not the same. Understanding why will help you to make better use of a calculator and to write correct mathematical equations.

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    $\begingroup$ Which pocket calculator are you using? Mine gives $13$ because typing $1\cdot 0$ gives $1.0$ which is just$~1$. $\endgroup$ – Marc van Leeuwen May 29 '13 at 12:19
  • $\begingroup$ @MarcvanLeeuwen - good spot: I was assuming "dot" to be "times" of course - an easy assumption to make, and perhaps, as you suggest, a mistake! $\endgroup$ – Mark Bennet May 29 '13 at 12:24
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\begin{align*} & 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + (1\times0) + 1\\ = &1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 1\\ = &12 \end{align*}

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$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$

since priority of multiplication $\times$ or $\cdot$ is greater than addition $+$ so expression will be:

$1+1+1+1+1+1+1+1+1+1+1+0+1\implies 12$

This is the order of operation:

$1$ B:- Brackets first

$2$ O:- Orders (i.e. Powers and Square Roots, etc.)

$3$ DM:- Division and Multiplication (left-to-right)

$4$ AS:- Addition and Subtraction (left-to-right)

I think this will helpful :

http://www.magicalmaths.org/bodmas-starter-5-5-5-5-5-5-5-5-x-0-i-bet-more-people-will-answer-it-wrongp-give-a-try/

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$1+1+1+1+1+1+1+1+1+1+1+1⋅0+1$

So, $1*0=0$ witch means that we Are Left with

$1+1+1+1+1+1+1+1+1+1+1+1(+0)$ witch is $12$

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That would equal $12$. That is because, according to PEMDAS, you would have to solve the multiplication first. That would turn your problem into $1+1+1+1+1+1+1+1+1+1+1+(1×0)+1$. That turns it into $1+1+1+1+1+1+1+1+1+1+1+0+1$, and that would equal $12$, so the final answer would be be $12$.

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  • $\begingroup$ When you answer a seven year old question that already has many answers essentially identical to yours you waste your time and, since the answer bumps the question to the active list, the time of people who scan that list for new information. Please don't do that again. $\endgroup$ – Ethan Bolker Jul 27 at 2:00
  • $\begingroup$ Okay, I'm deeply sorry. I won't do that again. $\endgroup$ – Mister Pro 2 days ago
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Use the PEMDAS law: $1+1+1+1+1+1+1+1+1+1+1+1×0+1=1+1+1+1+1+1+1+1+1+1+1+0+1=1+1+1+1+1+1+1+1+1+1+1+1=12$ Hopes that is helpful

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