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.
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Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
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.
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$endgroup$
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
17k31434
$endgroup$
$begingroup$
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
1 hour ago
add a comment |
Your Answer
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2 Answers
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$begingroup$
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
17k31434
$endgroup$
$begingroup$
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
1 hour 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
17k31434
$endgroup$
$begingroup$
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
1 hour 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
17k31434
$endgroup$
$$beginalign
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
endalign$$
answered 1 hour ago
saulspatzsaulspatz
17k31434
answered 1 hour ago
saulspatzsaulspatz
17k31434
answered 1 hour ago
saulspatzsaulspatz
17k31434
answered 1 hour ago
saulspatzsaulspatz
17k31434
17k31434
$begingroup$
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
1 hour 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
1 hour 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
1 hour 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
1 hour 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.
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$begingroup$
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