$1+1+1+1+1+1+1+1+1+1+1+1 cdot 0+1 = 12$ or $1$?
$begingroup$
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.
arithmetic
edited May 29 '13 at 11:21
Sharkos
13.4k22350
asked May 29 '13 at 8:51
Dev01Dev01
1293
$endgroup$
migrated from mathematica.stackexchange.com May 29 '13 at 8:57
This question came from our site for users of Wolfram Mathematica.
|
show 8 more comments
$begingroup$
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.
arithmetic
edited May 29 '13 at 11:21
Sharkos
13.4k22350
asked May 29 '13 at 8:51
Dev01Dev01
1293
$endgroup$
migrated from mathematica.stackexchange.com May 29 '13 at 8:57
This question came from our site for users of Wolfram Mathematica.
1
$begingroup$
linear-algebra? modular-arithmetic? Why did someone edit those tags in??
$endgroup$
– mrf
May 29 '13 at 9:03
2
$begingroup$
@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
3
$begingroup$
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.
$endgroup$
– Sharkos
May 29 '13 at 9:05
2
$begingroup$
@Sharkos "Dear" downvoters? : )
$endgroup$
– Rudy the Reindeer
May 29 '13 at 9:06
4
$begingroup$
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.
$endgroup$
– Sharkos
May 29 '13 at 9:11
|
show 8 more comments
$begingroup$
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.
arithmetic
edited May 29 '13 at 11:21
Sharkos
13.4k22350
asked May 29 '13 at 8:51
Dev01Dev01
1293
$endgroup$
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.
arithmetic
arithmetic
edited May 29 '13 at 11:21
Sharkos
13.4k22350
asked May 29 '13 at 8:51
Dev01Dev01
1293
edited May 29 '13 at 11:21
Sharkos
13.4k22350
asked May 29 '13 at 8:51
Dev01Dev01
1293
edited May 29 '13 at 11:21
Sharkos
13.4k22350
edited May 29 '13 at 11:21
Sharkos
13.4k22350
edited May 29 '13 at 11:21
Sharkos
13.4k22350
13.4k22350
asked May 29 '13 at 8:51
Dev01Dev01
1293
asked May 29 '13 at 8:51
Dev01Dev01
1293
asked May 29 '13 at 8:51
Dev01Dev01
1293
1293
migrated from mathematica.stackexchange.com May 29 '13 at 8:57
This question came from our site for users of Wolfram Mathematica.
migrated from mathematica.stackexchange.com May 29 '13 at 8:57
This question came from our site for users of Wolfram Mathematica.
1
$begingroup$
linear-algebra? modular-arithmetic? Why did someone edit those tags in??
$endgroup$
– mrf
May 29 '13 at 9:03
2
$begingroup$
@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
3
$begingroup$
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.
$endgroup$
– Sharkos
May 29 '13 at 9:05
2
$begingroup$
@Sharkos "Dear" downvoters? : )
$endgroup$
– Rudy the Reindeer
May 29 '13 at 9:06
4
$begingroup$
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.
$endgroup$
– Sharkos
May 29 '13 at 9:11
|
show 8 more comments
1
$begingroup$
linear-algebra? modular-arithmetic? Why did someone edit those tags in??
$endgroup$
– mrf
May 29 '13 at 9:03
2
$begingroup$
@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
3
$begingroup$
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.
$endgroup$
– Sharkos
May 29 '13 at 9:05
2
$begingroup$
@Sharkos "Dear" downvoters? : )
$endgroup$
– Rudy the Reindeer
May 29 '13 at 9:06
4
$begingroup$
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.
$endgroup$
– Sharkos
May 29 '13 at 9:11
1
1
$begingroup$
linear-algebra? modular-arithmetic? Why did someone edit those tags in??
$endgroup$
– mrf
May 29 '13 at 9:03
$begingroup$
linear-algebra? modular-arithmetic? Why did someone edit those tags in??
$endgroup$
– mrf
May 29 '13 at 9:03
2
2
$begingroup$
@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
$begingroup$
@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
3
3
$begingroup$
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.
$endgroup$
– Sharkos
May 29 '13 at 9:05
$begingroup$
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.
$endgroup$
– Sharkos
May 29 '13 at 9:05
2
2
$begingroup$
@Sharkos "Dear" downvoters? : )
$endgroup$
– Rudy the Reindeer
May 29 '13 at 9:06
$begingroup$
@Sharkos "Dear" downvoters? : )
$endgroup$
– Rudy the Reindeer
May 29 '13 at 9:06
4
4
$begingroup$
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.
$endgroup$
– Sharkos
May 29 '13 at 9:11
$begingroup$
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.
$endgroup$
– Sharkos
May 29 '13 at 9:11
|
show 8 more comments
6 Answers
6
active
oldest
votes
$begingroup$
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
$$1+1+1+1+1+1+1+1+1+1+1+1cdot 0+1$$
is to be evaluated as
$$1+1+1+1+1+1+1+1+1+1+1+(1cdot 0)+1;,$$
not as
$$(1+1+1+1+1+1+1+1+1+1+1+1)cdot(0+1);.$$
Since $1cdot0=0$, this simplifies to
$$1+1+1+1+1+1+1+1+1+1+1+0+1=12;.$$
answered May 29 '13 at 9:08
Brian M. ScottBrian M. Scott
460k40516917
$endgroup$
add a comment |
$begingroup$
Multiplication takes priority over addition, so this
$1+1+1+1+1+1+1+1+1+1+1+1times0+1$
becomes:
$1+1+1+1+1+1+1+1+1+1+1+0+1$
Now add everything which gives you 12.
answered May 29 '13 at 8:59
JerryJerry
831615
$endgroup$
add a comment |
$begingroup$
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.
answered May 29 '13 at 9:17
Mark BennetMark Bennet
81.9k984183
$endgroup$
1
$begingroup$
Which pocket calculator are you using? Mine gives $13$ because typing $1cdot 0$ 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
add a comment |
$begingroup$
begin{align*}
& 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + (1times0) + 1\
= &1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 1\
= &12
end{align*}
edited May 29 '13 at 9:18
Michael Albanese
64.4k1599314
answered May 29 '13 at 9:00
user54297user54297
426517
$endgroup$
add a comment |
$begingroup$
$1+1+1+1+1+1+1+1+1+1+1+1 cdot 0+1$
since priority of multiplication $times$ or $cdot$ is greater than addition $+$ so expression will be:
$1+1+1+1+1+1+1+1+1+1+1+0+1implies 12$
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/
answered May 29 '13 at 9:12
iostream007iostream007
3,71931439
$endgroup$
add a comment |
$begingroup$
$1+1+1+1+1+1+1+1+1+1+1+1⋅0+1$
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 $12$
answered Dec 12 '18 at 1:15
A.usernemaA.usernema
113
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$1+1+1+1+1+1+1+1+1+1+1+1cdot 0+1$$
is to be evaluated as
$$1+1+1+1+1+1+1+1+1+1+1+(1cdot 0)+1;,$$
not as
$$(1+1+1+1+1+1+1+1+1+1+1+1)cdot(0+1);.$$
Since $1cdot0=0$, this simplifies to
$$1+1+1+1+1+1+1+1+1+1+1+0+1=12;.$$
answered May 29 '13 at 9:08
Brian M. ScottBrian M. Scott
460k40516917
$endgroup$
add a comment |
$begingroup$
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
$$1+1+1+1+1+1+1+1+1+1+1+1cdot 0+1$$
is to be evaluated as
$$1+1+1+1+1+1+1+1+1+1+1+(1cdot 0)+1;,$$
not as
$$(1+1+1+1+1+1+1+1+1+1+1+1)cdot(0+1);.$$
Since $1cdot0=0$, this simplifies to
$$1+1+1+1+1+1+1+1+1+1+1+0+1=12;.$$
answered May 29 '13 at 9:08
Brian M. ScottBrian M. Scott
460k40516917
$endgroup$
add a comment |
$begingroup$
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
$$1+1+1+1+1+1+1+1+1+1+1+1cdot 0+1$$
is to be evaluated as
$$1+1+1+1+1+1+1+1+1+1+1+(1cdot 0)+1;,$$
not as
$$(1+1+1+1+1+1+1+1+1+1+1+1)cdot(0+1);.$$
Since $1cdot0=0$, this simplifies to
$$1+1+1+1+1+1+1+1+1+1+1+0+1=12;.$$
answered May 29 '13 at 9:08
Brian M. ScottBrian M. Scott
460k40516917
$endgroup$
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
$$1+1+1+1+1+1+1+1+1+1+1+1cdot 0+1$$
is to be evaluated as
$$1+1+1+1+1+1+1+1+1+1+1+(1cdot 0)+1;,$$
not as
$$(1+1+1+1+1+1+1+1+1+1+1+1)cdot(0+1);.$$
Since $1cdot0=0$, this simplifies to
$$1+1+1+1+1+1+1+1+1+1+1+0+1=12;.$$
answered May 29 '13 at 9:08
Brian M. ScottBrian M. Scott
460k40516917
answered May 29 '13 at 9:08
Brian M. ScottBrian M. Scott
460k40516917
answered May 29 '13 at 9:08
Brian M. ScottBrian M. Scott
460k40516917
answered May 29 '13 at 9:08
Brian M. ScottBrian M. Scott
460k40516917
460k40516917
add a comment |
add a comment |
$begingroup$
Multiplication takes priority over addition, so this
$1+1+1+1+1+1+1+1+1+1+1+1times0+1$
becomes:
$1+1+1+1+1+1+1+1+1+1+1+0+1$
Now add everything which gives you 12.
answered May 29 '13 at 8:59
JerryJerry
831615
$endgroup$
add a comment |
$begingroup$
Multiplication takes priority over addition, so this
$1+1+1+1+1+1+1+1+1+1+1+1times0+1$
becomes:
$1+1+1+1+1+1+1+1+1+1+1+0+1$
Now add everything which gives you 12.
answered May 29 '13 at 8:59
JerryJerry
831615
$endgroup$
add a comment |
$begingroup$
Multiplication takes priority over addition, so this
$1+1+1+1+1+1+1+1+1+1+1+1times0+1$
becomes:
$1+1+1+1+1+1+1+1+1+1+1+0+1$
Now add everything which gives you 12.
answered May 29 '13 at 8:59
JerryJerry
831615
$endgroup$
Multiplication takes priority over addition, so this
$1+1+1+1+1+1+1+1+1+1+1+1times0+1$
becomes:
$1+1+1+1+1+1+1+1+1+1+1+0+1$
Now add everything which gives you 12.
answered May 29 '13 at 8:59
JerryJerry
831615
answered May 29 '13 at 8:59
JerryJerry
831615
answered May 29 '13 at 8:59
JerryJerry
831615
answered May 29 '13 at 8:59
JerryJerry
831615
831615
add a comment |
add a comment |
$begingroup$
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.
answered May 29 '13 at 9:17
Mark BennetMark Bennet
81.9k984183
$endgroup$
1
$begingroup$
Which pocket calculator are you using? Mine gives $13$ because typing $1cdot 0$ 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
add a comment |
$begingroup$
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.
answered May 29 '13 at 9:17
Mark BennetMark Bennet
81.9k984183
$endgroup$
1
$begingroup$
Which pocket calculator are you using? Mine gives $13$ because typing $1cdot 0$ 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
add a comment |
$begingroup$
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.
answered May 29 '13 at 9:17
Mark BennetMark Bennet
81.9k984183
$endgroup$
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.
answered May 29 '13 at 9:17
Mark BennetMark Bennet
81.9k984183
answered May 29 '13 at 9:17
Mark BennetMark Bennet
81.9k984183
answered May 29 '13 at 9:17
Mark BennetMark Bennet
81.9k984183
answered May 29 '13 at 9:17
Mark BennetMark Bennet
81.9k984183
81.9k984183
1
$begingroup$
Which pocket calculator are you using? Mine gives $13$ because typing $1cdot 0$ 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
add a comment |
1
$begingroup$
Which pocket calculator are you using? Mine gives $13$ because typing $1cdot 0$ 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
1
1
$begingroup$
Which pocket calculator are you using? Mine gives $13$ because typing $1cdot 0$ gives $1.0$ which is just$~1$.
$endgroup$
– Marc van Leeuwen
May 29 '13 at 12:19
$begingroup$
Which pocket calculator are you using? Mine gives $13$ because typing $1cdot 0$ 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
$begingroup$
@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
add a comment |
$begingroup$
begin{align*}
& 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + (1times0) + 1\
= &1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 1\
= &12
end{align*}
edited May 29 '13 at 9:18
Michael Albanese
64.4k1599314
answered May 29 '13 at 9:00
user54297user54297
426517
$endgroup$
add a comment |
$begingroup$
begin{align*}
& 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + (1times0) + 1\
= &1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 1\
= &12
end{align*}
edited May 29 '13 at 9:18
Michael Albanese
64.4k1599314
answered May 29 '13 at 9:00
user54297user54297
426517
$endgroup$
add a comment |
$begingroup$
begin{align*}
& 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + (1times0) + 1\
= &1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 1\
= &12
end{align*}
edited May 29 '13 at 9:18
Michael Albanese
64.4k1599314
answered May 29 '13 at 9:00
user54297user54297
426517
$endgroup$
begin{align*}
& 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + (1times0) + 1\
= &1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 1\
= &12
end{align*}
edited May 29 '13 at 9:18
Michael Albanese
64.4k1599314
answered May 29 '13 at 9:00
user54297user54297
426517
edited May 29 '13 at 9:18
Michael Albanese
64.4k1599314
edited May 29 '13 at 9:18
Michael Albanese
64.4k1599314
edited May 29 '13 at 9:18
Michael Albanese
64.4k1599314
64.4k1599314
answered May 29 '13 at 9:00
user54297user54297
426517
answered May 29 '13 at 9:00
user54297user54297
426517
answered May 29 '13 at 9:00
user54297user54297
426517
426517
add a comment |
add a comment |
$begingroup$
$1+1+1+1+1+1+1+1+1+1+1+1 cdot 0+1$
since priority of multiplication $times$ or $cdot$ is greater than addition $+$ so expression will be:
$1+1+1+1+1+1+1+1+1+1+1+0+1implies 12$
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/
answered May 29 '13 at 9:12
iostream007iostream007
3,71931439
$endgroup$
add a comment |
$begingroup$
$1+1+1+1+1+1+1+1+1+1+1+1 cdot 0+1$
since priority of multiplication $times$ or $cdot$ is greater than addition $+$ so expression will be:
$1+1+1+1+1+1+1+1+1+1+1+0+1implies 12$
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/
answered May 29 '13 at 9:12
iostream007iostream007
3,71931439
$endgroup$
add a comment |
$begingroup$
$1+1+1+1+1+1+1+1+1+1+1+1 cdot 0+1$
since priority of multiplication $times$ or $cdot$ is greater than addition $+$ so expression will be:
$1+1+1+1+1+1+1+1+1+1+1+0+1implies 12$
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/
answered May 29 '13 at 9:12
iostream007iostream007
3,71931439
$endgroup$
$1+1+1+1+1+1+1+1+1+1+1+1 cdot 0+1$
since priority of multiplication $times$ or $cdot$ is greater than addition $+$ so expression will be:
$1+1+1+1+1+1+1+1+1+1+1+0+1implies 12$
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/
answered May 29 '13 at 9:12
iostream007iostream007
3,71931439
edited May 29 '13 at 9:25
answered May 29 '13 at 9:12
iostream007iostream007
3,71931439
answered May 29 '13 at 9:12
iostream007iostream007
3,71931439
answered May 29 '13 at 9:12
iostream007iostream007
3,71931439
3,71931439
add a comment |
add a comment |
$begingroup$
$1+1+1+1+1+1+1+1+1+1+1+1⋅0+1$
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 $12$
answered Dec 12 '18 at 1:15
A.usernemaA.usernema
113
$endgroup$
add a comment |
$begingroup$
$1+1+1+1+1+1+1+1+1+1+1+1⋅0+1$
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 $12$
answered Dec 12 '18 at 1:15
A.usernemaA.usernema
113
$endgroup$
add a comment |
$begingroup$
$1+1+1+1+1+1+1+1+1+1+1+1⋅0+1$
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 $12$
answered Dec 12 '18 at 1:15
A.usernemaA.usernema
113
$endgroup$
$1+1+1+1+1+1+1+1+1+1+1+1⋅0+1$
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 $12$
answered Dec 12 '18 at 1:15
A.usernemaA.usernema
113
answered Dec 12 '18 at 1:15
A.usernemaA.usernema
113
answered Dec 12 '18 at 1:15
A.usernemaA.usernema
113
answered Dec 12 '18 at 1:15
A.usernemaA.usernema
113
113
add a comment |
add a comment |
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1
$begingroup$
linear-algebra? modular-arithmetic? Why did someone edit those tags in??
$endgroup$
– mrf
May 29 '13 at 9:03
2
$begingroup$
@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
3
$begingroup$
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.
$endgroup$
– Sharkos
May 29 '13 at 9:05
2
$begingroup$
@Sharkos "Dear" downvoters? : )
$endgroup$
– Rudy the Reindeer
May 29 '13 at 9:06
4
$begingroup$
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.
$endgroup$
– Sharkos
May 29 '13 at 9:11