Is it possible to find 2014 distinct positive integers whose sum is divisible by each of...
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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
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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 |
$begingroup$
$$begin{align}
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
end{align}$$
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
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 |
$begingroup$
$$begin{align}
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
end{align}$$
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
1 hour ago
add a comment |
$begingroup$
$$begin{align}
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
end{align}$$
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
1 hour ago
add a comment |
$begingroup$
$$begin{align}
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
end{align}$$
answered 1 hour ago
saulspatzsaulspatz
16.9k31434
$endgroup$
$$begin{align}
1+2+3&=6\
1+2+3+6&=12\
1+2+3+6+12&=24\
vdots
end{align}$$
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
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