Prove that in a row of 17 random integers there exist some integers written in succession (next to each other), whose sum is divisible by 17. [duplicate]In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?Let $S=3,4,5,6,7,8,9,10,11,12$. Suppose 6 integers are chosen from S. Must there be 2 integers whose sum is 15?Prove that if $|S| ge 2^n−1 + 1$, then $S$ contains two elements which are disjoint from each other.Let S be a set of n integers. Show that there is a subset of S, sum of whose elements is a multiple of n by pigeon holeProve that any $6$- subset of integers $1…14$ is always the union of two distinct subsets of equal sumProve that given any five integers, there will be three for which the sum of the squares of those integers is divisible by 3.
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Prove that in a row of 17 random integers there exist some integers written in succession (next to each other), whose sum is divisible by 17. [duplicate]
In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?Let $S=3,4,5,6,7,8,9,10,11,12$. Suppose 6 integers are chosen from S. Must there be 2 integers whose sum is 15?Prove that if $|S| ge 2^n−1 + 1$, then $S$ contains two elements which are disjoint from each other.Let S be a set of n integers. Show that there is a subset of S, sum of whose elements is a multiple of n by pigeon holeProve that any $6$- subset of integers $1…14$ is always the union of two distinct subsets of equal sumProve that given any five integers, there will be three for which the sum of the squares of those integers is divisible by 3.
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This question already has an answer here:
In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?
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I presume I have to use the pigeonhole principle here, but so far no luck.
discrete-mathematics
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marked as duplicate by MJD, Javi, Shailesh, nmasanta, Feng Shao 2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?
1 answer
I presume I have to use the pigeonhole principle here, but so far no luck.
discrete-mathematics
New contributor
$endgroup$
marked as duplicate by MJD, Javi, Shailesh, nmasanta, Feng Shao 2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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@Dzoooks The OP says that the integers are random.
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– saulspatz
10 hours ago
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@Dzoooks You are allowed to take more than two integers. Or only one. Why do you think it isn't true if they are random?
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– saulspatz
10 hours ago
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@Dzoooks Read the first sentence of my last comment again. Or look at my answer.
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– saulspatz
10 hours ago
add a comment
|
$begingroup$
This question already has an answer here:
In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?
1 answer
I presume I have to use the pigeonhole principle here, but so far no luck.
discrete-mathematics
New contributor
$endgroup$
This question already has an answer here:
In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?
1 answer
I presume I have to use the pigeonhole principle here, but so far no luck.
This question already has an answer here:
In a sequence of $n$ integers, must there be a contiguous subsequence that sums to a multiple of $n$?
1 answer
discrete-mathematics
discrete-mathematics
New contributor
New contributor
edited 9 hours ago
N. F. Taussig
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asked 10 hours ago
BiobbbBiobbb
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marked as duplicate by MJD, Javi, Shailesh, nmasanta, Feng Shao 2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by MJD, Javi, Shailesh, nmasanta, Feng Shao 2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by MJD, Javi, Shailesh, nmasanta, Feng Shao 2 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
@Dzoooks The OP says that the integers are random.
$endgroup$
– saulspatz
10 hours ago
$begingroup$
@Dzoooks You are allowed to take more than two integers. Or only one. Why do you think it isn't true if they are random?
$endgroup$
– saulspatz
10 hours ago
$begingroup$
@Dzoooks Read the first sentence of my last comment again. Or look at my answer.
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– saulspatz
10 hours ago
add a comment
|
1
$begingroup$
@Dzoooks The OP says that the integers are random.
$endgroup$
– saulspatz
10 hours ago
$begingroup$
@Dzoooks You are allowed to take more than two integers. Or only one. Why do you think it isn't true if they are random?
$endgroup$
– saulspatz
10 hours ago
$begingroup$
@Dzoooks Read the first sentence of my last comment again. Or look at my answer.
$endgroup$
– saulspatz
10 hours ago
1
1
$begingroup$
@Dzoooks The OP says that the integers are random.
$endgroup$
– saulspatz
10 hours ago
$begingroup$
@Dzoooks The OP says that the integers are random.
$endgroup$
– saulspatz
10 hours ago
$begingroup$
@Dzoooks You are allowed to take more than two integers. Or only one. Why do you think it isn't true if they are random?
$endgroup$
– saulspatz
10 hours ago
$begingroup$
@Dzoooks You are allowed to take more than two integers. Or only one. Why do you think it isn't true if they are random?
$endgroup$
– saulspatz
10 hours ago
$begingroup$
@Dzoooks Read the first sentence of my last comment again. Or look at my answer.
$endgroup$
– saulspatz
10 hours ago
$begingroup$
@Dzoooks Read the first sentence of my last comment again. Or look at my answer.
$endgroup$
– saulspatz
10 hours ago
add a comment
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2 Answers
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Consider $$a_1,a_1+a_2,a_1+a_2+a_3,...,a_1+a_2+...a_17$$
If remainders in dividing by $17$ are different one has $0$ remainder otherwise two of them have the same remainders. In both cases the problem is solved.
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HINTS
Let the numbers be $a_1,a_2,dots,a_17$ and let $s_k=sum_i=1^ka_i$ for $k=1,dots,17.$ If $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider $$a_1,a_1+a_2,a_1+a_2+a_3,...,a_1+a_2+...a_17$$
If remainders in dividing by $17$ are different one has $0$ remainder otherwise two of them have the same remainders. In both cases the problem is solved.
$endgroup$
add a comment
|
$begingroup$
Consider $$a_1,a_1+a_2,a_1+a_2+a_3,...,a_1+a_2+...a_17$$
If remainders in dividing by $17$ are different one has $0$ remainder otherwise two of them have the same remainders. In both cases the problem is solved.
$endgroup$
add a comment
|
$begingroup$
Consider $$a_1,a_1+a_2,a_1+a_2+a_3,...,a_1+a_2+...a_17$$
If remainders in dividing by $17$ are different one has $0$ remainder otherwise two of them have the same remainders. In both cases the problem is solved.
$endgroup$
Consider $$a_1,a_1+a_2,a_1+a_2+a_3,...,a_1+a_2+...a_17$$
If remainders in dividing by $17$ are different one has $0$ remainder otherwise two of them have the same remainders. In both cases the problem is solved.
edited 10 hours ago
answered 10 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
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54.7k4 gold badges27 silver badges74 bronze badges
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add a comment
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$begingroup$
HINTS
Let the numbers be $a_1,a_2,dots,a_17$ and let $s_k=sum_i=1^ka_i$ for $k=1,dots,17.$ If $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.
$endgroup$
add a comment
|
$begingroup$
HINTS
Let the numbers be $a_1,a_2,dots,a_17$ and let $s_k=sum_i=1^ka_i$ for $k=1,dots,17.$ If $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.
$endgroup$
add a comment
|
$begingroup$
HINTS
Let the numbers be $a_1,a_2,dots,a_17$ and let $s_k=sum_i=1^ka_i$ for $k=1,dots,17.$ If $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.
$endgroup$
HINTS
Let the numbers be $a_1,a_2,dots,a_17$ and let $s_k=sum_i=1^ka_i$ for $k=1,dots,17.$ If $s_jequiv s_kpmod17$ for some $j<k$ what can you conclude? Now finish it off with the pigeonhole principle.
edited 9 hours ago
MJD
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49k31 gold badges219 silver badges408 bronze badges
answered 10 hours ago
saulspatzsaulspatz
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25.4k4 gold badges16 silver badges42 bronze badges
add a comment
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add a comment
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1
$begingroup$
@Dzoooks The OP says that the integers are random.
$endgroup$
– saulspatz
10 hours ago
$begingroup$
@Dzoooks You are allowed to take more than two integers. Or only one. Why do you think it isn't true if they are random?
$endgroup$
– saulspatz
10 hours ago
$begingroup$
@Dzoooks Read the first sentence of my last comment again. Or look at my answer.
$endgroup$
– saulspatz
10 hours ago