# $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$?

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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## migrated from mathematica.stackexchange.comMay 29 '13 at 8:57

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linear-algebra? modular-arithmetic? Why did someone edit those tags in?? – mrf May 29 '13 at 9:03
@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. – Brian M. Scott May 29 '13 at 9:04
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. – Sharkos May 29 '13 at 9:05
@Sharkos "Dear" downvoters? : ) – Rudy the Reindeer May 29 '13 at 9:06
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. – Sharkos May 29 '13 at 9:11

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

\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 :