Mixing
If $12mid{a}$ then $4nmid{b}$ or $4mid{(8+2a+5b)}$.
$begingroup$
Prove that if $12mid{a}$ then $4nmid{b}$ or $4mid{(8+2a+5b)}$.
I have gotten thus far:
Since $12mid{a}$, then $a=12k$ for some integer $k$.
Substitute $a=12k$ into $(8+2a+5b)$ to get $8+24k+5b$.
Not sure what to do from here, any help is appreciated, thanks!
elementary-number-theory proof-writing divisibility
edited Jan 27 at 23:02
Martin Sleziak
44.9k10122276
asked Jan 27 at 22:21
macymacy
526
$endgroup$
|
show 3 more comments
$begingroup$
Prove that if $12mid{a}$ then $4nmid{b}$ or $4mid{(8+2a+5b)}$.
I have gotten thus far:
Since $12mid{a}$, then $a=12k$ for some integer $k$.
Substitute $a=12k$ into $(8+2a+5b)$ to get $8+24k+5b$.
Not sure what to do from here, any help is appreciated, thanks!
elementary-number-theory proof-writing divisibility
edited Jan 27 at 23:02
Martin Sleziak
44.9k10122276
asked Jan 27 at 22:21
macymacy
526
$endgroup$
4
$begingroup$
Hint: if $4 ,nmid ,b$ then you are done. So, now assume $b=4m$
$endgroup$
– lulu
Jan 27 at 22:23
$begingroup$
@lulu Why are you answering in comments? (Hints are answers)
$endgroup$
– Arthur
Jan 27 at 22:30
$begingroup$
$large 2mid a,Rightarrow,4mid 2a,,$ so $ large 4mid 8!+!2a!+!5biff 4mid 5biff 4mid b $
$endgroup$
– Bill Dubuque
Jan 28 at 1:08
$begingroup$
But isn't it that $4nmid{b}$?
$endgroup$
– macy
Jan 28 at 3:07
1
$begingroup$
@Peter It's not about reputation. It's about getting an answer out there and getting the system to understand that (for instance getting this off the unanswered queue, and make this a viable target for closing other questions as duplicates). This site as a whole would've been (margnally) better off if lulu had put that hint in an answer post.
$endgroup$
– Arthur
Jan 28 at 11:49
|
show 3 more comments
$begingroup$
Prove that if $12mid{a}$ then $4nmid{b}$ or $4mid{(8+2a+5b)}$.
I have gotten thus far:
Since $12mid{a}$, then $a=12k$ for some integer $k$.
Substitute $a=12k$ into $(8+2a+5b)$ to get $8+24k+5b$.
Not sure what to do from here, any help is appreciated, thanks!
elementary-number-theory proof-writing divisibility
edited Jan 27 at 23:02
Martin Sleziak
44.9k10122276
asked Jan 27 at 22:21
macymacy
526
$endgroup$
Prove that if $12mid{a}$ then $4nmid{b}$ or $4mid{(8+2a+5b)}$.
I have gotten thus far:
Since $12mid{a}$, then $a=12k$ for some integer $k$.
Substitute $a=12k$ into $(8+2a+5b)$ to get $8+24k+5b$.
Not sure what to do from here, any help is appreciated, thanks!
elementary-number-theory proof-writing divisibility
elementary-number-theory proof-writing divisibility
edited Jan 27 at 23:02
Martin Sleziak
44.9k10122276
asked Jan 27 at 22:21
macymacy
526
edited Jan 27 at 23:02
Martin Sleziak
44.9k10122276
asked Jan 27 at 22:21
macymacy
526
edited Jan 27 at 23:02
Martin Sleziak
44.9k10122276
edited Jan 27 at 23:02
Martin Sleziak
44.9k10122276
edited Jan 27 at 23:02
Martin Sleziak
44.9k10122276
44.9k10122276
asked Jan 27 at 22:21
macymacy
526
asked Jan 27 at 22:21
macymacy
526
asked Jan 27 at 22:21
macymacy
526
526
4
$begingroup$
Hint: if $4 ,nmid ,b$ then you are done. So, now assume $b=4m$
$endgroup$
– lulu
Jan 27 at 22:23
$begingroup$
@lulu Why are you answering in comments? (Hints are answers)
$endgroup$
– Arthur
Jan 27 at 22:30
$begingroup$
$large 2mid a,Rightarrow,4mid 2a,,$ so $ large 4mid 8!+!2a!+!5biff 4mid 5biff 4mid b $
$endgroup$
– Bill Dubuque
Jan 28 at 1:08
$begingroup$
But isn't it that $4nmid{b}$?
$endgroup$
– macy
Jan 28 at 3:07
1
$begingroup$
@Peter It's not about reputation. It's about getting an answer out there and getting the system to understand that (for instance getting this off the unanswered queue, and make this a viable target for closing other questions as duplicates). This site as a whole would've been (margnally) better off if lulu had put that hint in an answer post.
$endgroup$
– Arthur
Jan 28 at 11:49
|
show 3 more comments
4
$begingroup$
Hint: if $4 ,nmid ,b$ then you are done. So, now assume $b=4m$
$endgroup$
– lulu
Jan 27 at 22:23
$begingroup$
@lulu Why are you answering in comments? (Hints are answers)
$endgroup$
– Arthur
Jan 27 at 22:30
$begingroup$
$large 2mid a,Rightarrow,4mid 2a,,$ so $ large 4mid 8!+!2a!+!5biff 4mid 5biff 4mid b $
$endgroup$
– Bill Dubuque
Jan 28 at 1:08
$begingroup$
But isn't it that $4nmid{b}$?
$endgroup$
– macy
Jan 28 at 3:07
1
$begingroup$
@Peter It's not about reputation. It's about getting an answer out there and getting the system to understand that (for instance getting this off the unanswered queue, and make this a viable target for closing other questions as duplicates). This site as a whole would've been (margnally) better off if lulu had put that hint in an answer post.
$endgroup$
– Arthur
Jan 28 at 11:49
4
4
$begingroup$
Hint: if $4 ,nmid ,b$ then you are done. So, now assume $b=4m$
$endgroup$
– lulu
Jan 27 at 22:23
$begingroup$
Hint: if $4 ,nmid ,b$ then you are done. So, now assume $b=4m$
$endgroup$
– lulu
Jan 27 at 22:23
$begingroup$
@lulu Why are you answering in comments? (Hints are answers)
$endgroup$
– Arthur
Jan 27 at 22:30
$begingroup$
@lulu Why are you answering in comments? (Hints are answers)
$endgroup$
– Arthur
Jan 27 at 22:30
$begingroup$
$large 2mid a,Rightarrow,4mid 2a,,$ so $ large 4mid 8!+!2a!+!5biff 4mid 5biff 4mid b $
$endgroup$
– Bill Dubuque
Jan 28 at 1:08
$begingroup$
$large 2mid a,Rightarrow,4mid 2a,,$ so $ large 4mid 8!+!2a!+!5biff 4mid 5biff 4mid b $
$endgroup$
– Bill Dubuque
Jan 28 at 1:08
$begingroup$
But isn't it that $4nmid{b}$?
$endgroup$
– macy
Jan 28 at 3:07
$begingroup$
But isn't it that $4nmid{b}$?
$endgroup$
– macy
Jan 28 at 3:07
1
1
$begingroup$
@Peter It's not about reputation. It's about getting an answer out there and getting the system to understand that (for instance getting this off the unanswered queue, and make this a viable target for closing other questions as duplicates). This site as a whole would've been (margnally) better off if lulu had put that hint in an answer post.
$endgroup$
– Arthur
Jan 28 at 11:49
$begingroup$
@Peter It's not about reputation. It's about getting an answer out there and getting the system to understand that (for instance getting this off the unanswered queue, and make this a viable target for closing other questions as duplicates). This site as a whole would've been (margnally) better off if lulu had put that hint in an answer post.
$endgroup$
– Arthur
Jan 28 at 11:49
|
show 3 more comments
1 Answer
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$begingroup$
Assume $12|a$ and $4|b$ , otherwise nothing has to be proven, then , we can write $a=4s$ and $b=4t$ with intgegers $s,t$
Then, we have $$8+2a+5b=8+8s+20t=4(2+2s+5t)$$ hence $4$ divides $8+2a+5b$
answered Jan 28 at 11:54
PeterPeter
49.1k1240136
$endgroup$
add a comment |
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$begingroup$
Assume $12|a$ and $4|b$ , otherwise nothing has to be proven, then , we can write $a=4s$ and $b=4t$ with intgegers $s,t$
Then, we have $$8+2a+5b=8+8s+20t=4(2+2s+5t)$$ hence $4$ divides $8+2a+5b$
answered Jan 28 at 11:54
PeterPeter
49.1k1240136
$endgroup$
add a comment |
$begingroup$
Assume $12|a$ and $4|b$ , otherwise nothing has to be proven, then , we can write $a=4s$ and $b=4t$ with intgegers $s,t$
Then, we have $$8+2a+5b=8+8s+20t=4(2+2s+5t)$$ hence $4$ divides $8+2a+5b$
answered Jan 28 at 11:54
PeterPeter
49.1k1240136
$endgroup$
add a comment |
$begingroup$
Assume $12|a$ and $4|b$ , otherwise nothing has to be proven, then , we can write $a=4s$ and $b=4t$ with intgegers $s,t$
Then, we have $$8+2a+5b=8+8s+20t=4(2+2s+5t)$$ hence $4$ divides $8+2a+5b$
answered Jan 28 at 11:54
PeterPeter
49.1k1240136
$endgroup$
Assume $12|a$ and $4|b$ , otherwise nothing has to be proven, then , we can write $a=4s$ and $b=4t$ with intgegers $s,t$
Then, we have $$8+2a+5b=8+8s+20t=4(2+2s+5t)$$ hence $4$ divides $8+2a+5b$
answered Jan 28 at 11:54
PeterPeter
49.1k1240136
answered Jan 28 at 11:54
PeterPeter
49.1k1240136
answered Jan 28 at 11:54
PeterPeter
49.1k1240136
answered Jan 28 at 11:54
PeterPeter
49.1k1240136
49.1k1240136
add a comment |
add a comment |
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4
$begingroup$
Hint: if $4 ,nmid ,b$ then you are done. So, now assume $b=4m$
$endgroup$
– lulu
Jan 27 at 22:23
$begingroup$
@lulu Why are you answering in comments? (Hints are answers)
$endgroup$
– Arthur
Jan 27 at 22:30
$begingroup$
$large 2mid a,Rightarrow,4mid 2a,,$ so $ large 4mid 8!+!2a!+!5biff 4mid 5biff 4mid b $
$endgroup$
– Bill Dubuque
Jan 28 at 1:08
$begingroup$
But isn't it that $4nmid{b}$?
$endgroup$
– macy
Jan 28 at 3:07
1
$begingroup$
@Peter It's not about reputation. It's about getting an answer out there and getting the system to understand that (for instance getting this off the unanswered queue, and make this a viable target for closing other questions as duplicates). This site as a whole would've been (margnally) better off if lulu had put that hint in an answer post.
$endgroup$
– Arthur
Jan 28 at 11:49