Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?Permutations_Combination DiscreteProve that there must be two distinct integers in $A$ whose sum is $104$.Pigeon Hole Principle: Six positive integers whose maximum is at most $14$Find the number of possible combinations for a combination lock if each combination…Exhibit a one-to-one correspondence between the set of positive integers and the set of integers not divisible by $3$.Is it possible to find two distinct 4-colorings of the tetrahedron which use exactly one of each color?Among any $11$ integers, sum of $6$ of them is divisible by $6$Prove that any collection of 8 distinct integers contains distinct x and y such that x - y is divisible by 7.Show that given a set of positive n integers, there exists a non-empty subset whose sum is divisible by nPigeonhole Principle Issue five integers where their sum or difference is divisible by seven.
Does an unused member variable take up memory?
QQ Plot and Shapiro Wilk Test Disagree
Plausibility of Mushroom Buildings
What is the population of Romulus in the TNG era?
Trig Subsitution When There's No Square Root
Is it a Cyclops number? "Nobody" knows!
Are small insurances worth it?
Gaining more land
What are you allowed to do while using the Warlock's Eldritch Master feature?
ER diagram relationship node size adjustment
Does "Until when" sound natural for native speakers?
Possible to detect presence of nuclear bomb?
How does Ehrenfest's theorem apply to the quantum harmonic oscillator?
How exactly does an Ethernet collision happen in the cable, since nodes use different circuits for Tx and Rx?
Can the alpha, lambda values of a glmnet object output determine whether ridge or Lasso?
Can one live in the U.S. and not use a credit card?
Confusion about Complex Continued Fraction
What do *foreign films* mean for an American?
What is the generally accepted pronunciation of “topoi”?
When Schnorr signatures are part of Bitcoin will it be possible validate each block with only one signature validation?
How do electrons receive energy when a body is heated?
Can't make sense of a paragraph from Lovecraft
How do spaceships determine each other's mass in space?
Why is a very small peak with larger m/z not considered to be the molecular ion?
Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
Permutations_Combination DiscreteProve that there must be two distinct integers in $A$ whose sum is $104$.Pigeon Hole Principle: Six positive integers whose maximum is at most $14$Find the number of possible combinations for a combination lock if each combination…Exhibit a one-to-one correspondence between the set of positive integers and the set of integers not divisible by $3$.Is it possible to find two distinct 4-colorings of the tetrahedron which use exactly one of each color?Among any $11$ integers, sum of $6$ of them is divisible by $6$Prove that any collection of 8 distinct integers contains distinct x and y such that x - y is divisible by 7.Show that given a set of positive n integers, there exists a non-empty subset whose sum is divisible by nPigeonhole Principle Issue five integers where their sum or difference is divisible by seven.
$begingroup$
Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
I'm not really sure how to even approach this question.
Source: Washington's Monthly Math Hour, 2014
discrete-mathematics intuition pigeonhole-principle
asked 1 hour ago
Arvin DingArvin Ding
83
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
I'm not really sure how to even approach this question.
Source: Washington's Monthly Math Hour, 2014
discrete-mathematics intuition pigeonhole-principle
asked 1 hour ago
Arvin DingArvin Ding
83
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
1 hour ago
1
$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
1 hour ago
$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
1 hour ago
$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
1 hour ago
add a comment |
$begingroup$
Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
I'm not really sure how to even approach this question.
Source: Washington's Monthly Math Hour, 2014
discrete-mathematics intuition pigeonhole-principle
asked 1 hour ago
Arvin DingArvin Ding
83
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
I'm not really sure how to even approach this question.
Source: Washington's Monthly Math Hour, 2014
discrete-mathematics intuition pigeonhole-principle
discrete-mathematics intuition pigeonhole-principle
asked 1 hour ago
Arvin DingArvin Ding
83
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Arvin DingArvin Ding
83
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Arvin DingArvin Ding
83
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Arvin DingArvin Ding
83
asked 1 hour ago
Arvin DingArvin Ding
83
83
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Arvin Ding is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
1 hour ago
1
$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
1 hour ago
$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
1 hour ago
$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
1 hour ago
add a comment |
$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
1 hour ago
1
$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
1 hour ago
$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
1 hour ago
$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
1 hour ago
$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
1 hour ago
$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
1 hour ago
1
1
$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
1 hour ago
$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
1 hour ago
$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
1 hour ago
$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
1 hour ago
$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
1 hour ago
$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$
So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$
$~$
Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_n+1 = sumlimits_k=1^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_k=1^nx_k$ for all $ileq n$ for all $ngeq 3$.
answered 1 hour ago
JMoravitzJMoravitz
48.2k33886
$endgroup$
add a comment |
$begingroup$
$$beginalign
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
endalign$$
answered 1 hour ago
saulspatzsaulspatz
16.9k31434
$endgroup$
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
59 mins ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
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
);
);
Arvin Ding is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3143024%2fis-it-possible-to-find-2014-distinct-positive-integers-whose-sum-is-divisible-by%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$
So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$
$~$
Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_n+1 = sumlimits_k=1^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_k=1^nx_k$ for all $ileq n$ for all $ngeq 3$.
answered 1 hour ago
JMoravitzJMoravitz
48.2k33886
$endgroup$
add a comment |
$begingroup$
Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$
So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$
$~$
Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_n+1 = sumlimits_k=1^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_k=1^nx_k$ for all $ileq n$ for all $ngeq 3$.
answered 1 hour ago
JMoravitzJMoravitz
48.2k33886
$endgroup$
add a comment |
$begingroup$
Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$
So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$
$~$
Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_n+1 = sumlimits_k=1^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_k=1^nx_k$ for all $ileq n$ for all $ngeq 3$.
answered 1 hour ago
JMoravitzJMoravitz
48.2k33886
$endgroup$
Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$
So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$
$~$
Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_n+1 = sumlimits_k=1^n x_k$ for each $ngeq 3$. You have that $x_imid sumlimits_k=1^nx_k$ for all $ileq n$ for all $ngeq 3$.
answered 1 hour ago
JMoravitzJMoravitz
48.2k33886
answered 1 hour ago
JMoravitzJMoravitz
48.2k33886
answered 1 hour ago
JMoravitzJMoravitz
48.2k33886
answered 1 hour ago
JMoravitzJMoravitz
48.2k33886
48.2k33886
add a comment |
add a comment |
$begingroup$
$$beginalign
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
endalign$$
answered 1 hour ago
saulspatzsaulspatz
16.9k31434
$endgroup$
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
59 mins ago
add a comment |
$begingroup$
$$beginalign
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
endalign$$
answered 1 hour ago
saulspatzsaulspatz
16.9k31434
$endgroup$
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
59 mins ago
add a comment |
$begingroup$
$$beginalign
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
endalign$$
answered 1 hour ago
saulspatzsaulspatz
16.9k31434
$endgroup$
$$beginalign
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
endalign$$
answered 1 hour ago
saulspatzsaulspatz
16.9k31434
answered 1 hour ago
saulspatzsaulspatz
16.9k31434
answered 1 hour ago
saulspatzsaulspatz
16.9k31434
answered 1 hour ago
saulspatzsaulspatz
16.9k31434
16.9k31434
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
59 mins ago
add a comment |
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
59 mins ago
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
59 mins ago
$begingroup$
We can say that the $2014$ numbers produced this way are $1,2,3$ and $3cdot 2^k$ for $k in [1,2011]$
$endgroup$
– Ross Millikan
59 mins ago
add a comment |
Arvin Ding is a new contributor. Be nice, and check out our Code of Conduct.
Arvin Ding is a new contributor. Be nice, and check out our Code of Conduct.
Arvin Ding is a new contributor. Be nice, and check out our Code of Conduct.
Arvin Ding is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3143024%2fis-it-possible-to-find-2014-distinct-positive-integers-whose-sum-is-divisible-by%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
$begingroup$
I cant even find two.
$endgroup$
– Rudi_Birnbaum
1 hour ago
1
$begingroup$
@Rudi_Birnbaum irrelevant. $2,4,6,12$ are all divisors of $2+4+6+12$.
$endgroup$
– JMoravitz
1 hour ago
$begingroup$
Well, at least I can find three… $1+2+3$
$endgroup$
– Wolfgang Kais
1 hour ago
$begingroup$
@JMoravitz humor?
$endgroup$
– Rudi_Birnbaum
1 hour ago