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Why is it that the product $(a-x)(b-x) cdots(y-x)(z-x)$ simplifies to $0$? [on hold]
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$begingroup$
I came across this trick question, and I don't understand the solution to the puzzle. We are asked to simplify the product
$$(a-x)(b-x) cdots(y-x)(z-x),tag{$*$}$$
and the answer is apparently supposed to be $0$.
I have no idea how this can be, and how one can simplify the above expression. Can someone explain to me why the product $(*)$
simplifies to $0$?
algebra-precalculus puzzle
edited 23 hours ago
Brahadeesh
6,47942363
asked Oct 9 '14 at 4:53
user182015user182015
71
$endgroup$
put on hold as off-topic by Saad, José Carlos Santos, stressed out, Xander Henderson, ancientmathematician 16 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, José Carlos Santos, stressed out, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I came across this trick question, and I don't understand the solution to the puzzle. We are asked to simplify the product
$$(a-x)(b-x) cdots(y-x)(z-x),tag{$*$}$$
and the answer is apparently supposed to be $0$.
I have no idea how this can be, and how one can simplify the above expression. Can someone explain to me why the product $(*)$
simplifies to $0$?
algebra-precalculus puzzle
edited 23 hours ago
Brahadeesh
6,47942363
asked Oct 9 '14 at 4:53
user182015user182015
71
$endgroup$
put on hold as off-topic by Saad, José Carlos Santos, stressed out, Xander Henderson, ancientmathematician 16 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, José Carlos Santos, stressed out, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
16
$begingroup$
It has a factor $x-x$
$endgroup$
– lab bhattacharjee
Oct 9 '14 at 4:54
add a comment |
$begingroup$
I came across this trick question, and I don't understand the solution to the puzzle. We are asked to simplify the product
$$(a-x)(b-x) cdots(y-x)(z-x),tag{$*$}$$
and the answer is apparently supposed to be $0$.
I have no idea how this can be, and how one can simplify the above expression. Can someone explain to me why the product $(*)$
simplifies to $0$?
algebra-precalculus puzzle
edited 23 hours ago
Brahadeesh
6,47942363
asked Oct 9 '14 at 4:53
user182015user182015
71
$endgroup$
I came across this trick question, and I don't understand the solution to the puzzle. We are asked to simplify the product
$$(a-x)(b-x) cdots(y-x)(z-x),tag{$*$}$$
and the answer is apparently supposed to be $0$.
I have no idea how this can be, and how one can simplify the above expression. Can someone explain to me why the product $(*)$
simplifies to $0$?
algebra-precalculus puzzle
algebra-precalculus puzzle
edited 23 hours ago
Brahadeesh
6,47942363
asked Oct 9 '14 at 4:53
user182015user182015
71
edited 23 hours ago
Brahadeesh
6,47942363
asked Oct 9 '14 at 4:53
user182015user182015
71
edited 23 hours ago
Brahadeesh
6,47942363
edited 23 hours ago
Brahadeesh
6,47942363
edited 23 hours ago
Brahadeesh
6,47942363
6,47942363
asked Oct 9 '14 at 4:53
user182015user182015
71
asked Oct 9 '14 at 4:53
user182015user182015
71
asked Oct 9 '14 at 4:53
user182015user182015
71
71
put on hold as off-topic by Saad, José Carlos Santos, stressed out, Xander Henderson, ancientmathematician 16 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, José Carlos Santos, stressed out, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Saad, José Carlos Santos, stressed out, Xander Henderson, ancientmathematician 16 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, José Carlos Santos, stressed out, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
16
$begingroup$
It has a factor $x-x$
$endgroup$
– lab bhattacharjee
Oct 9 '14 at 4:54
add a comment |
16
$begingroup$
It has a factor $x-x$
$endgroup$
– lab bhattacharjee
Oct 9 '14 at 4:54
16
16
$begingroup$
It has a factor $x-x$
$endgroup$
– lab bhattacharjee
Oct 9 '14 at 4:54
$begingroup$
It has a factor $x-x$
$endgroup$
– lab bhattacharjee
Oct 9 '14 at 4:54
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Its a puzzle and a trick question and not a legitimate math question.
We are tricked into think that there is a variable we call $x$ and a sequence of constants $a,b,c,....,y,z$ and we need to find the product of $(a-x)(b-x)... (y-x)(z-x)$ and that is hard. But if we use the naming conventions of letters then one of our constants will be $x$ and our variable, which we intuitively think of as a different animal altogether, is actually a constant. So one of our terms is $(x-x)=0$ and the product is $0$.
But this is nonsense.
Every letter is a variable and we have no idea what they represent so we do not know what values they will be so we can not use $....$ to represent going from $a$ to $b$ to $c$ etc so we can't know that the unnamed terms are $(c-x), (d-x),(e-x)$ etc. This is simply badly written math.
If we legitimately wanted to do something like this we'd have to index the variables in some way perhaps like $(a_1 -x)(a_2 - x) ..... (a_{26} - x)$ with it understood there is a known or calculable sequence of variable $a_1, a_2, a_3....$.
If we wrote this as $(a_1 - a_{24})(a_2-a_{24})(a_3 - a_{24}).... (a_{25} - a_{24})(a_{26} - a_{24})$ it should be obvious that one of the terms is $(a_{24}-a_{24}) = 0$.
answered yesterday
fleabloodfleablood
72k22687
$endgroup$
add a comment |
$begingroup$
As lab bhattacharjee pointed out (assuming the implied pattern is to be followed), one of the terms in that expression (namely, the term before the third shown term) is $x-x$, which is of course equal to $0$ and causes the whole expression to be equal to $0$ as well.
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Its a puzzle and a trick question and not a legitimate math question.
We are tricked into think that there is a variable we call $x$ and a sequence of constants $a,b,c,....,y,z$ and we need to find the product of $(a-x)(b-x)... (y-x)(z-x)$ and that is hard. But if we use the naming conventions of letters then one of our constants will be $x$ and our variable, which we intuitively think of as a different animal altogether, is actually a constant. So one of our terms is $(x-x)=0$ and the product is $0$.
But this is nonsense.
Every letter is a variable and we have no idea what they represent so we do not know what values they will be so we can not use $....$ to represent going from $a$ to $b$ to $c$ etc so we can't know that the unnamed terms are $(c-x), (d-x),(e-x)$ etc. This is simply badly written math.
If we legitimately wanted to do something like this we'd have to index the variables in some way perhaps like $(a_1 -x)(a_2 - x) ..... (a_{26} - x)$ with it understood there is a known or calculable sequence of variable $a_1, a_2, a_3....$.
If we wrote this as $(a_1 - a_{24})(a_2-a_{24})(a_3 - a_{24}).... (a_{25} - a_{24})(a_{26} - a_{24})$ it should be obvious that one of the terms is $(a_{24}-a_{24}) = 0$.
answered yesterday
fleabloodfleablood
72k22687
$endgroup$
add a comment |
$begingroup$
Its a puzzle and a trick question and not a legitimate math question.
We are tricked into think that there is a variable we call $x$ and a sequence of constants $a,b,c,....,y,z$ and we need to find the product of $(a-x)(b-x)... (y-x)(z-x)$ and that is hard. But if we use the naming conventions of letters then one of our constants will be $x$ and our variable, which we intuitively think of as a different animal altogether, is actually a constant. So one of our terms is $(x-x)=0$ and the product is $0$.
But this is nonsense.
Every letter is a variable and we have no idea what they represent so we do not know what values they will be so we can not use $....$ to represent going from $a$ to $b$ to $c$ etc so we can't know that the unnamed terms are $(c-x), (d-x),(e-x)$ etc. This is simply badly written math.
If we legitimately wanted to do something like this we'd have to index the variables in some way perhaps like $(a_1 -x)(a_2 - x) ..... (a_{26} - x)$ with it understood there is a known or calculable sequence of variable $a_1, a_2, a_3....$.
If we wrote this as $(a_1 - a_{24})(a_2-a_{24})(a_3 - a_{24}).... (a_{25} - a_{24})(a_{26} - a_{24})$ it should be obvious that one of the terms is $(a_{24}-a_{24}) = 0$.
answered yesterday
fleabloodfleablood
72k22687
$endgroup$
add a comment |
$begingroup$
Its a puzzle and a trick question and not a legitimate math question.
We are tricked into think that there is a variable we call $x$ and a sequence of constants $a,b,c,....,y,z$ and we need to find the product of $(a-x)(b-x)... (y-x)(z-x)$ and that is hard. But if we use the naming conventions of letters then one of our constants will be $x$ and our variable, which we intuitively think of as a different animal altogether, is actually a constant. So one of our terms is $(x-x)=0$ and the product is $0$.
But this is nonsense.
Every letter is a variable and we have no idea what they represent so we do not know what values they will be so we can not use $....$ to represent going from $a$ to $b$ to $c$ etc so we can't know that the unnamed terms are $(c-x), (d-x),(e-x)$ etc. This is simply badly written math.
If we legitimately wanted to do something like this we'd have to index the variables in some way perhaps like $(a_1 -x)(a_2 - x) ..... (a_{26} - x)$ with it understood there is a known or calculable sequence of variable $a_1, a_2, a_3....$.
If we wrote this as $(a_1 - a_{24})(a_2-a_{24})(a_3 - a_{24}).... (a_{25} - a_{24})(a_{26} - a_{24})$ it should be obvious that one of the terms is $(a_{24}-a_{24}) = 0$.
answered yesterday
fleabloodfleablood
72k22687
$endgroup$
Its a puzzle and a trick question and not a legitimate math question.
We are tricked into think that there is a variable we call $x$ and a sequence of constants $a,b,c,....,y,z$ and we need to find the product of $(a-x)(b-x)... (y-x)(z-x)$ and that is hard. But if we use the naming conventions of letters then one of our constants will be $x$ and our variable, which we intuitively think of as a different animal altogether, is actually a constant. So one of our terms is $(x-x)=0$ and the product is $0$.
But this is nonsense.
Every letter is a variable and we have no idea what they represent so we do not know what values they will be so we can not use $....$ to represent going from $a$ to $b$ to $c$ etc so we can't know that the unnamed terms are $(c-x), (d-x),(e-x)$ etc. This is simply badly written math.
If we legitimately wanted to do something like this we'd have to index the variables in some way perhaps like $(a_1 -x)(a_2 - x) ..... (a_{26} - x)$ with it understood there is a known or calculable sequence of variable $a_1, a_2, a_3....$.
If we wrote this as $(a_1 - a_{24})(a_2-a_{24})(a_3 - a_{24}).... (a_{25} - a_{24})(a_{26} - a_{24})$ it should be obvious that one of the terms is $(a_{24}-a_{24}) = 0$.
answered yesterday
fleabloodfleablood
72k22687
answered yesterday
fleabloodfleablood
72k22687
answered yesterday
fleabloodfleablood
72k22687
answered yesterday
fleabloodfleablood
72k22687
72k22687
add a comment |
add a comment |
$begingroup$
As lab bhattacharjee pointed out (assuming the implied pattern is to be followed), one of the terms in that expression (namely, the term before the third shown term) is $x-x$, which is of course equal to $0$ and causes the whole expression to be equal to $0$ as well.
$endgroup$
add a comment |
$begingroup$
As lab bhattacharjee pointed out (assuming the implied pattern is to be followed), one of the terms in that expression (namely, the term before the third shown term) is $x-x$, which is of course equal to $0$ and causes the whole expression to be equal to $0$ as well.
$endgroup$
add a comment |
$begingroup$
As lab bhattacharjee pointed out (assuming the implied pattern is to be followed), one of the terms in that expression (namely, the term before the third shown term) is $x-x$, which is of course equal to $0$ and causes the whole expression to be equal to $0$ as well.
$endgroup$
As lab bhattacharjee pointed out (assuming the implied pattern is to be followed), one of the terms in that expression (namely, the term before the third shown term) is $x-x$, which is of course equal to $0$ and causes the whole expression to be equal to $0$ as well.
answered yesterday
community wiki
Robert Howard
add a comment |
add a comment |
16
$begingroup$
It has a factor $x-x$
$endgroup$
– lab bhattacharjee
Oct 9 '14 at 4:54