In $mathbb Z_{10}$ show that $7mid5$ and in $mathbb Z_{12}$ show that $5mid8$ [closed]
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Help I don't know where to start. I don't think I understand if I'm in mod 10 it will never be $7$ away from $10$ only ever $5$.
abstract-algebra
edited Dec 19 '18 at 6:13
Saad
20k92352
asked Dec 19 '18 at 5:49
Wyatt EcklundWyatt Ecklund
6
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closed as off-topic by Brahadeesh, Stahl, Andrés E. Caicedo, KReiser, Chinnapparaj R Dec 19 '18 at 7:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
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$begingroup$
Help I don't know where to start. I don't think I understand if I'm in mod 10 it will never be $7$ away from $10$ only ever $5$.
abstract-algebra
edited Dec 19 '18 at 6:13
Saad
20k92352
asked Dec 19 '18 at 5:49
Wyatt EcklundWyatt Ecklund
6
$endgroup$
closed as off-topic by Brahadeesh, Stahl, Andrés E. Caicedo, KReiser, Chinnapparaj R Dec 19 '18 at 7:34
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." – Brahadeesh, KReiser, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
But $7 +7=2*7 equiv 4 pmod {10}$ so $7|4$. That's $4$ away; not $7$. And $7+ 7 + 7 = 3*7 equiv 1 pmod{10}$ and that's $1$ away; not $7$.
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– fleablood
Dec 19 '18 at 6:23
add a comment |
$begingroup$
Help I don't know where to start. I don't think I understand if I'm in mod 10 it will never be $7$ away from $10$ only ever $5$.
abstract-algebra
edited Dec 19 '18 at 6:13
Saad
20k92352
asked Dec 19 '18 at 5:49
Wyatt EcklundWyatt Ecklund
6
$endgroup$
Help I don't know where to start. I don't think I understand if I'm in mod 10 it will never be $7$ away from $10$ only ever $5$.
abstract-algebra
abstract-algebra
edited Dec 19 '18 at 6:13
Saad
20k92352
asked Dec 19 '18 at 5:49
Wyatt EcklundWyatt Ecklund
6
edited Dec 19 '18 at 6:13
Saad
20k92352
asked Dec 19 '18 at 5:49
Wyatt EcklundWyatt Ecklund
6
edited Dec 19 '18 at 6:13
Saad
20k92352
edited Dec 19 '18 at 6:13
Saad
20k92352
edited Dec 19 '18 at 6:13
Saad
20k92352
20k92352
asked Dec 19 '18 at 5:49
Wyatt EcklundWyatt Ecklund
6
asked Dec 19 '18 at 5:49
Wyatt EcklundWyatt Ecklund
6
asked Dec 19 '18 at 5:49
Wyatt EcklundWyatt Ecklund
6
6
closed as off-topic by Brahadeesh, Stahl, Andrés E. Caicedo, KReiser, Chinnapparaj R Dec 19 '18 at 7:34
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." – Brahadeesh, KReiser, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Brahadeesh, Stahl, Andrés E. Caicedo, KReiser, Chinnapparaj R Dec 19 '18 at 7:34
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." – Brahadeesh, KReiser, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
But $7 +7=2*7 equiv 4 pmod {10}$ so $7|4$. That's $4$ away; not $7$. And $7+ 7 + 7 = 3*7 equiv 1 pmod{10}$ and that's $1$ away; not $7$.
$endgroup$
– fleablood
Dec 19 '18 at 6:23
add a comment |
$begingroup$
But $7 +7=2*7 equiv 4 pmod {10}$ so $7|4$. That's $4$ away; not $7$. And $7+ 7 + 7 = 3*7 equiv 1 pmod{10}$ and that's $1$ away; not $7$.
$endgroup$
– fleablood
Dec 19 '18 at 6:23
$begingroup$
But $7 +7=2*7 equiv 4 pmod {10}$ so $7|4$. That's $4$ away; not $7$. And $7+ 7 + 7 = 3*7 equiv 1 pmod{10}$ and that's $1$ away; not $7$.
$endgroup$
– fleablood
Dec 19 '18 at 6:23
$begingroup$
But $7 +7=2*7 equiv 4 pmod {10}$ so $7|4$. That's $4$ away; not $7$. And $7+ 7 + 7 = 3*7 equiv 1 pmod{10}$ and that's $1$ away; not $7$.
$endgroup$
– fleablood
Dec 19 '18 at 6:23
add a comment |
1 Answer
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$begingroup$
If $7|5$, then there is some number $x$ such that $7xequiv 5pmod{10}$. Absent any intuition or insight, one can just try plugging in $x=0,1,2,ldots,9$. For each, simplify $7x$, modulo $10$, and see if you get $5$.
A similar strategy will solve the second problem.
answered Dec 19 '18 at 6:35
vadim123vadim123
76.4k897191
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $7|5$, then there is some number $x$ such that $7xequiv 5pmod{10}$. Absent any intuition or insight, one can just try plugging in $x=0,1,2,ldots,9$. For each, simplify $7x$, modulo $10$, and see if you get $5$.
A similar strategy will solve the second problem.
answered Dec 19 '18 at 6:35
vadim123vadim123
76.4k897191
$endgroup$
add a comment |
$begingroup$
If $7|5$, then there is some number $x$ such that $7xequiv 5pmod{10}$. Absent any intuition or insight, one can just try plugging in $x=0,1,2,ldots,9$. For each, simplify $7x$, modulo $10$, and see if you get $5$.
A similar strategy will solve the second problem.
answered Dec 19 '18 at 6:35
vadim123vadim123
76.4k897191
$endgroup$
add a comment |
$begingroup$
If $7|5$, then there is some number $x$ such that $7xequiv 5pmod{10}$. Absent any intuition or insight, one can just try plugging in $x=0,1,2,ldots,9$. For each, simplify $7x$, modulo $10$, and see if you get $5$.
A similar strategy will solve the second problem.
answered Dec 19 '18 at 6:35
vadim123vadim123
76.4k897191
$endgroup$
If $7|5$, then there is some number $x$ such that $7xequiv 5pmod{10}$. Absent any intuition or insight, one can just try plugging in $x=0,1,2,ldots,9$. For each, simplify $7x$, modulo $10$, and see if you get $5$.
A similar strategy will solve the second problem.
answered Dec 19 '18 at 6:35
vadim123vadim123
76.4k897191
answered Dec 19 '18 at 6:35
vadim123vadim123
76.4k897191
answered Dec 19 '18 at 6:35
vadim123vadim123
76.4k897191
answered Dec 19 '18 at 6:35
vadim123vadim123
76.4k897191
76.4k897191
add a comment |
add a comment |
$begingroup$
But $7 +7=2*7 equiv 4 pmod {10}$ so $7|4$. That's $4$ away; not $7$. And $7+ 7 + 7 = 3*7 equiv 1 pmod{10}$ and that's $1$ away; not $7$.
$endgroup$
– fleablood
Dec 19 '18 at 6:23