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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.

not all wrong
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Dev01
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  • @BrianM.Scott What's with all the downvotes? I don't get it. – Rudy the Reindeer May 29 '13 at 09:02
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    linear-algebra? modular-arithmetic? Why did someone edit those tags in?? – mrf May 29 '13 at 09:03
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    @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 09:04
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    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. – not all wrong May 29 '13 at 09:05
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    @Sharkos "Dear" downvoters? : ) – Rudy the Reindeer May 29 '13 at 09:06
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    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. – not all wrong May 29 '13 at 09:11
  • @Sharkos I am happy to inform you that you are "royally screwed" : ) – Rudy the Reindeer May 29 '13 at 09:17
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    @MattN,@Sharkos, having just seen the question I downvoted it, not because the OP was wrong, but because it's not the kind of question that I would like to see on the site. I don't think it's off topic enough to close, as I can see why people would want it here, but I don't like it so I downvoted. No offence meant to the asker! – Tom Oldfield May 29 '13 at 09:58
  • @TomOldfield Very nice of you, thank you for your comment. – Rudy the Reindeer May 29 '13 at 10:07
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    @TomOldfield Thanks for commenting; communication helps. I'd probably point out that the FAQ says nothing about questions being "too elementary" whatsoever, and indeed references http://meta.math.stackexchange.com/questions/1868/list-of-generalizations-of-common-questions which has an order of operations question itself. I don't see any reason to discourage people from asking this sort of question. Is there/could there be a tag for elementary questions? – not all wrong May 29 '13 at 11:06
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    To everybody: Unearthed an arithmetic tag, which I've substituted for the pre-calc one since it's more suitable. If questions on arithmetic are objectionable, I suggest taking this up on meta. – not all wrong May 29 '13 at 11:23
  • is 1+1+1+1+1+1+1+1+1+1+1 really needed at the beginning? just removing that would show some understanding of meaning of +. – jimjim May 29 '13 at 11:57
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    I didn't downvote, but here are some in my eyes valid criticisms of the question: (1) A question should try to focus on the essential; simplifying to "is $11+1\cdot 0+1=12$?" or even "is $n+1\cdot0+1=n+1$?" would have avoided possible confusion and useless counting by readers; (2) the example is badly chosen; the fact that $(n+1)\cdot(0+1)=n+1=n+(1\cdot0)+1$ makes that a "yes" answer does not tell which variant was used; (3) there is an unexplained use of a non-word "pemdas". – Marc van Leeuwen May 29 '13 at 12:15

6 Answers6

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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

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

Brian M. Scott
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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.

Jerry
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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*}

Michael Albanese
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user54297
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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.

Mark Bennet
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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
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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 $12$

A.usernema
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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/

iostream007
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