Prove if n is an odd integer, then 3n is odd.If the square of an integer is odd, then the integer is oddIf $n^2+10$ is odd then $n$ is odd.Proof that an odd integer multiplied by $3$ and squared is always oddLet c be an integer which is not divisible by 3. Then the equation $x^3 − x = c$ has no integer solutions.Proving that if $n in mathbbZ$ and $n^2 − 6n + 5$ is even, then $n$ must be odd.Prove: if $x$ is even, then $x + 5$ is odd.Prove that $n^4-n^2$ is divisible by $8$ if $n$ is an odd positive integer.Proof: Given that $x$ is a positive integer, prove…Prove that,if $n$ is an integer, then $n^2 + 15n = 1$ is oddProve that $n! > n^3$ for every integer $n ge 6$ using induction
Have two feature in matrix command
Why would an invisible personal shield be necessary?
Does dual boot harm a laptop battery or reduce its life?
What is a good example for artistic ND filter applications?
Piece of chess engine, which accomplishes move generation
How to improve king safety
How well would the Moon protect the Earth from an Asteroid?
Do 3/8 (37.5%) of Quadratics Have No x-Intercepts?
Was the Psych theme song written for the show?
Is there an antonym (a complementary antonym) for "spicy" or "hot" regarding food (I DO NOT mean "seasoned", but "hot")?
Complexity of verifying optimality in (mixed) integer programming
Argand formula and more for quaternions?
What are the cons of stateless password generators?
Why did Windows 95 crash the whole system but newer Windows only crashed programs?
Move DB/LOG files - Detach/Attach - Problem
Why did I lose on time with 3 pawns vs Knight. Shouldn't it be a draw?
What Marvel character has this 'W' symbol?
If the Moon were impacted by a suitably sized meteor, how long would it take to impact the Earth?
How does a poisoned arrow combine with the spell Conjure Barrage?
Can Lightning Lure be used to knock out a creature like a magical Taser?
How can I kill my goat?
GNU sort stable sort when sort does not know sort order
How does the Thief's Fast Hands feature interact with mundane and magical shields?
Who said "one can be a powerful king with a very small sceptre"?
Prove if n is an odd integer, then 3n is odd.
If the square of an integer is odd, then the integer is oddIf $n^2+10$ is odd then $n$ is odd.Proof that an odd integer multiplied by $3$ and squared is always oddLet c be an integer which is not divisible by 3. Then the equation $x^3 − x = c$ has no integer solutions.Proving that if $n in mathbbZ$ and $n^2 − 6n + 5$ is even, then $n$ must be odd.Prove: if $x$ is even, then $x + 5$ is odd.Prove that $n^4-n^2$ is divisible by $8$ if $n$ is an odd positive integer.Proof: Given that $x$ is a positive integer, prove…Prove that,if $n$ is an integer, then $n^2 + 15n = 1$ is oddProve that $n! > n^3$ for every integer $n ge 6$ using induction
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
So far I have:
Assume $n$ is odd, $n = 2k+1$ for some integer $k$.
Thenbeginalign3n&= 3(2k+1)\
&= 6k + 3endalign
I'm unsure where to go from here.
elementary-number-theory
edited 9 hours ago
José Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
asked 9 hours ago
Amanda_CAmanda_C
236 bronze badges
$endgroup$
add a comment |
$begingroup$
So far I have:
Assume $n$ is odd, $n = 2k+1$ for some integer $k$.
Thenbeginalign3n&= 3(2k+1)\
&= 6k + 3endalign
I'm unsure where to go from here.
elementary-number-theory
edited 9 hours ago
José Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
asked 9 hours ago
Amanda_CAmanda_C
236 bronze badges
$endgroup$
$begingroup$
Can you find an integer $z$ such that your $3k$ is $3z+1$?
$endgroup$
– Ethan Bolker
9 hours ago
$begingroup$
$6k$ is even. Add $3$ to it gives....
$endgroup$
– David G. Stork
9 hours ago
add a comment |
$begingroup$
So far I have:
Assume $n$ is odd, $n = 2k+1$ for some integer $k$.
Thenbeginalign3n&= 3(2k+1)\
&= 6k + 3endalign
I'm unsure where to go from here.
elementary-number-theory
edited 9 hours ago
José Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
asked 9 hours ago
Amanda_CAmanda_C
236 bronze badges
$endgroup$
So far I have:
Assume $n$ is odd, $n = 2k+1$ for some integer $k$.
Thenbeginalign3n&= 3(2k+1)\
&= 6k + 3endalign
I'm unsure where to go from here.
elementary-number-theory
elementary-number-theory
edited 9 hours ago
José Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
asked 9 hours ago
Amanda_CAmanda_C
236 bronze badges
edited 9 hours ago
José Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
asked 9 hours ago
Amanda_CAmanda_C
236 bronze badges
edited 9 hours ago
José Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
edited 9 hours ago
José Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
edited 9 hours ago
José Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
204k25 gold badges159 silver badges280 bronze badges
asked 9 hours ago
Amanda_CAmanda_C
236 bronze badges
asked 9 hours ago
Amanda_CAmanda_C
236 bronze badges
asked 9 hours ago
Amanda_CAmanda_C
236 bronze badges
236 bronze badges
$begingroup$
Can you find an integer $z$ such that your $3k$ is $3z+1$?
$endgroup$
– Ethan Bolker
9 hours ago
$begingroup$
$6k$ is even. Add $3$ to it gives....
$endgroup$
– David G. Stork
9 hours ago
add a comment |
$begingroup$
Can you find an integer $z$ such that your $3k$ is $3z+1$?
$endgroup$
– Ethan Bolker
9 hours ago
$begingroup$
$6k$ is even. Add $3$ to it gives....
$endgroup$
– David G. Stork
9 hours ago
$begingroup$
Can you find an integer $z$ such that your $3k$ is $3z+1$?
$endgroup$
– Ethan Bolker
9 hours ago
$begingroup$
Can you find an integer $z$ such that your $3k$ is $3z+1$?
$endgroup$
– Ethan Bolker
9 hours ago
$begingroup$
$6k$ is even. Add $3$ to it gives....
$endgroup$
– David G. Stork
9 hours ago
$begingroup$
$6k$ is even. Add $3$ to it gives....
$endgroup$
– David G. Stork
9 hours ago
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Use the fact that $6k+3=2(3k+1)+1$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
$endgroup$
$begingroup$
How does the 2(3𝑘+1)+1 help?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
Because the odd integers are precisely those that can be written as $2N+1$.
$endgroup$
– José Carlos Santos
9 hours ago
$begingroup$
and so that means I can then write 2N + 1 where the N is actually the 3k + 1 ,,,which means its odd and the proof is true?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
A proof is neither true nor false. But if you add that $6k+3=2(2k+1)+1$, then it will be a correct proof.
$endgroup$
– José Carlos Santos
9 hours ago
add a comment |
$begingroup$
Since $2n$ is even, we have $3n=2n+n$ as a sum of an odd number and an even number, which is obviously odd.
answered 9 hours ago
szw1710szw1710
6,83712 silver badges23 bronze badges
$endgroup$
add a comment |
$begingroup$
Take $$n=2k+1$$ this is an odd number, then $$3(2k+1)=6k+3=(6k+2)+1$$ this is also an odd number.
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
86k4 gold badges29 silver badges69 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $n$ be an odd integer.
Then $3n$ is the product of two odd integers, and thus odd.
answered 7 hours ago
mlchristiansmlchristians
2,0852 silver badges19 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $n in mathbbZ$ be odd, so $n = 2m + 1$ for some $m in mathbbZ$. We have:
beginalign*
3n = 3(2m+1) = 6m + 3 = 2(3m + 1) + 1,
endalign*
where $3m + 1 in mathbbZ$ by closure. Let $z = 3m + 1$, so we have
beginalign*
3n = 2z + 1.
endalign*
Thus, $3n$ is odd.
answered 8 hours ago
Matt.PMatt.P
1,2015 silver badges19 bronze badges
$endgroup$
add a comment |
$begingroup$
$6k + 3$ can be written as $6k + 2 + 1$, now factoring we get $2(3k+1) + 1$, let $(3k+1) = A$, then $2A + 1$ is odd. in general, multiplying an odd integer with an odd integer always yields an odd integer. It is a good idea to prove it, you can also prove it with induction.
answered 8 hours ago
Donlans DonlansDonlans Donlans
275 bronze badges
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3308605%2fprove-if-n-is-an-odd-integer-then-3n-is-odd%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use the fact that $6k+3=2(3k+1)+1$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
$endgroup$
$begingroup$
How does the 2(3𝑘+1)+1 help?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
Because the odd integers are precisely those that can be written as $2N+1$.
$endgroup$
– José Carlos Santos
9 hours ago
$begingroup$
and so that means I can then write 2N + 1 where the N is actually the 3k + 1 ,,,which means its odd and the proof is true?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
A proof is neither true nor false. But if you add that $6k+3=2(2k+1)+1$, then it will be a correct proof.
$endgroup$
– José Carlos Santos
9 hours ago
add a comment |
$begingroup$
Use the fact that $6k+3=2(3k+1)+1$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
$endgroup$
$begingroup$
How does the 2(3𝑘+1)+1 help?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
Because the odd integers are precisely those that can be written as $2N+1$.
$endgroup$
– José Carlos Santos
9 hours ago
$begingroup$
and so that means I can then write 2N + 1 where the N is actually the 3k + 1 ,,,which means its odd and the proof is true?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
A proof is neither true nor false. But if you add that $6k+3=2(2k+1)+1$, then it will be a correct proof.
$endgroup$
– José Carlos Santos
9 hours ago
add a comment |
$begingroup$
Use the fact that $6k+3=2(3k+1)+1$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
$endgroup$
Use the fact that $6k+3=2(3k+1)+1$.
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
answered 9 hours ago
José Carlos SantosJosé Carlos Santos
204k25 gold badges159 silver badges280 bronze badges
204k25 gold badges159 silver badges280 bronze badges
$begingroup$
How does the 2(3𝑘+1)+1 help?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
Because the odd integers are precisely those that can be written as $2N+1$.
$endgroup$
– José Carlos Santos
9 hours ago
$begingroup$
and so that means I can then write 2N + 1 where the N is actually the 3k + 1 ,,,which means its odd and the proof is true?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
A proof is neither true nor false. But if you add that $6k+3=2(2k+1)+1$, then it will be a correct proof.
$endgroup$
– José Carlos Santos
9 hours ago
add a comment |
$begingroup$
How does the 2(3𝑘+1)+1 help?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
Because the odd integers are precisely those that can be written as $2N+1$.
$endgroup$
– José Carlos Santos
9 hours ago
$begingroup$
and so that means I can then write 2N + 1 where the N is actually the 3k + 1 ,,,which means its odd and the proof is true?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
A proof is neither true nor false. But if you add that $6k+3=2(2k+1)+1$, then it will be a correct proof.
$endgroup$
– José Carlos Santos
9 hours ago
$begingroup$
How does the 2(3𝑘+1)+1 help?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
How does the 2(3𝑘+1)+1 help?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
Because the odd integers are precisely those that can be written as $2N+1$.
$endgroup$
– José Carlos Santos
9 hours ago
$begingroup$
Because the odd integers are precisely those that can be written as $2N+1$.
$endgroup$
– José Carlos Santos
9 hours ago
$begingroup$
and so that means I can then write 2N + 1 where the N is actually the 3k + 1 ,,,which means its odd and the proof is true?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
and so that means I can then write 2N + 1 where the N is actually the 3k + 1 ,,,which means its odd and the proof is true?
$endgroup$
– Amanda_C
9 hours ago
$begingroup$
A proof is neither true nor false. But if you add that $6k+3=2(2k+1)+1$, then it will be a correct proof.
$endgroup$
– José Carlos Santos
9 hours ago
$begingroup$
A proof is neither true nor false. But if you add that $6k+3=2(2k+1)+1$, then it will be a correct proof.
$endgroup$
– José Carlos Santos
9 hours ago
add a comment |
$begingroup$
Since $2n$ is even, we have $3n=2n+n$ as a sum of an odd number and an even number, which is obviously odd.
answered 9 hours ago
szw1710szw1710
6,83712 silver badges23 bronze badges
$endgroup$
add a comment |
$begingroup$
Since $2n$ is even, we have $3n=2n+n$ as a sum of an odd number and an even number, which is obviously odd.
answered 9 hours ago
szw1710szw1710
6,83712 silver badges23 bronze badges
$endgroup$
add a comment |
$begingroup$
Since $2n$ is even, we have $3n=2n+n$ as a sum of an odd number and an even number, which is obviously odd.
answered 9 hours ago
szw1710szw1710
6,83712 silver badges23 bronze badges
$endgroup$
Since $2n$ is even, we have $3n=2n+n$ as a sum of an odd number and an even number, which is obviously odd.
answered 9 hours ago
szw1710szw1710
6,83712 silver badges23 bronze badges
answered 9 hours ago
szw1710szw1710
6,83712 silver badges23 bronze badges
answered 9 hours ago
szw1710szw1710
6,83712 silver badges23 bronze badges
answered 9 hours ago
szw1710szw1710
6,83712 silver badges23 bronze badges
6,83712 silver badges23 bronze badges
add a comment |
add a comment |
$begingroup$
Take $$n=2k+1$$ this is an odd number, then $$3(2k+1)=6k+3=(6k+2)+1$$ this is also an odd number.
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
86k4 gold badges29 silver badges69 bronze badges
$endgroup$
add a comment |
$begingroup$
Take $$n=2k+1$$ this is an odd number, then $$3(2k+1)=6k+3=(6k+2)+1$$ this is also an odd number.
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
86k4 gold badges29 silver badges69 bronze badges
$endgroup$
add a comment |
$begingroup$
Take $$n=2k+1$$ this is an odd number, then $$3(2k+1)=6k+3=(6k+2)+1$$ this is also an odd number.
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
86k4 gold badges29 silver badges69 bronze badges
$endgroup$
Take $$n=2k+1$$ this is an odd number, then $$3(2k+1)=6k+3=(6k+2)+1$$ this is also an odd number.
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
86k4 gold badges29 silver badges69 bronze badges
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
86k4 gold badges29 silver badges69 bronze badges
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
86k4 gold badges29 silver badges69 bronze badges
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
86k4 gold badges29 silver badges69 bronze badges
86k4 gold badges29 silver badges69 bronze badges
add a comment |
add a comment |
$begingroup$
Let $n$ be an odd integer.
Then $3n$ is the product of two odd integers, and thus odd.
answered 7 hours ago
mlchristiansmlchristians
2,0852 silver badges19 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $n$ be an odd integer.
Then $3n$ is the product of two odd integers, and thus odd.
answered 7 hours ago
mlchristiansmlchristians
2,0852 silver badges19 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $n$ be an odd integer.
Then $3n$ is the product of two odd integers, and thus odd.
answered 7 hours ago
mlchristiansmlchristians
2,0852 silver badges19 bronze badges
$endgroup$
Let $n$ be an odd integer.
Then $3n$ is the product of two odd integers, and thus odd.
answered 7 hours ago
mlchristiansmlchristians
2,0852 silver badges19 bronze badges
edited 3 hours ago
answered 7 hours ago
mlchristiansmlchristians
2,0852 silver badges19 bronze badges
answered 7 hours ago
mlchristiansmlchristians
2,0852 silver badges19 bronze badges
answered 7 hours ago
mlchristiansmlchristians
2,0852 silver badges19 bronze badges
2,0852 silver badges19 bronze badges
add a comment |
add a comment |
$begingroup$
Let $n in mathbbZ$ be odd, so $n = 2m + 1$ for some $m in mathbbZ$. We have:
beginalign*
3n = 3(2m+1) = 6m + 3 = 2(3m + 1) + 1,
endalign*
where $3m + 1 in mathbbZ$ by closure. Let $z = 3m + 1$, so we have
beginalign*
3n = 2z + 1.
endalign*
Thus, $3n$ is odd.
answered 8 hours ago
Matt.PMatt.P
1,2015 silver badges19 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $n in mathbbZ$ be odd, so $n = 2m + 1$ for some $m in mathbbZ$. We have:
beginalign*
3n = 3(2m+1) = 6m + 3 = 2(3m + 1) + 1,
endalign*
where $3m + 1 in mathbbZ$ by closure. Let $z = 3m + 1$, so we have
beginalign*
3n = 2z + 1.
endalign*
Thus, $3n$ is odd.
answered 8 hours ago
Matt.PMatt.P
1,2015 silver badges19 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $n in mathbbZ$ be odd, so $n = 2m + 1$ for some $m in mathbbZ$. We have:
beginalign*
3n = 3(2m+1) = 6m + 3 = 2(3m + 1) + 1,
endalign*
where $3m + 1 in mathbbZ$ by closure. Let $z = 3m + 1$, so we have
beginalign*
3n = 2z + 1.
endalign*
Thus, $3n$ is odd.
answered 8 hours ago
Matt.PMatt.P
1,2015 silver badges19 bronze badges
$endgroup$
Let $n in mathbbZ$ be odd, so $n = 2m + 1$ for some $m in mathbbZ$. We have:
beginalign*
3n = 3(2m+1) = 6m + 3 = 2(3m + 1) + 1,
endalign*
where $3m + 1 in mathbbZ$ by closure. Let $z = 3m + 1$, so we have
beginalign*
3n = 2z + 1.
endalign*
Thus, $3n$ is odd.
answered 8 hours ago
Matt.PMatt.P
1,2015 silver badges19 bronze badges
answered 8 hours ago
Matt.PMatt.P
1,2015 silver badges19 bronze badges
answered 8 hours ago
Matt.PMatt.P
1,2015 silver badges19 bronze badges
answered 8 hours ago
Matt.PMatt.P
1,2015 silver badges19 bronze badges
1,2015 silver badges19 bronze badges
add a comment |
add a comment |
$begingroup$
$6k + 3$ can be written as $6k + 2 + 1$, now factoring we get $2(3k+1) + 1$, let $(3k+1) = A$, then $2A + 1$ is odd. in general, multiplying an odd integer with an odd integer always yields an odd integer. It is a good idea to prove it, you can also prove it with induction.
answered 8 hours ago
Donlans DonlansDonlans Donlans
275 bronze badges
$endgroup$
add a comment |
$begingroup$
$6k + 3$ can be written as $6k + 2 + 1$, now factoring we get $2(3k+1) + 1$, let $(3k+1) = A$, then $2A + 1$ is odd. in general, multiplying an odd integer with an odd integer always yields an odd integer. It is a good idea to prove it, you can also prove it with induction.
answered 8 hours ago
Donlans DonlansDonlans Donlans
275 bronze badges
$endgroup$
add a comment |
$begingroup$
$6k + 3$ can be written as $6k + 2 + 1$, now factoring we get $2(3k+1) + 1$, let $(3k+1) = A$, then $2A + 1$ is odd. in general, multiplying an odd integer with an odd integer always yields an odd integer. It is a good idea to prove it, you can also prove it with induction.
answered 8 hours ago
Donlans DonlansDonlans Donlans
275 bronze badges
$endgroup$
$6k + 3$ can be written as $6k + 2 + 1$, now factoring we get $2(3k+1) + 1$, let $(3k+1) = A$, then $2A + 1$ is odd. in general, multiplying an odd integer with an odd integer always yields an odd integer. It is a good idea to prove it, you can also prove it with induction.
answered 8 hours ago
Donlans DonlansDonlans Donlans
275 bronze badges
answered 8 hours ago
Donlans DonlansDonlans Donlans
275 bronze badges
answered 8 hours ago
Donlans DonlansDonlans Donlans
275 bronze badges
answered 8 hours ago
Donlans DonlansDonlans Donlans
275 bronze badges
275 bronze badges
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3308605%2fprove-if-n-is-an-odd-integer-then-3n-is-odd%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Can you find an integer $z$ such that your $3k$ is $3z+1$?
$endgroup$
– Ethan Bolker
9 hours ago
$begingroup$
$6k$ is even. Add $3$ to it gives....
$endgroup$
– David G. Stork
9 hours ago