Let $S$ be a subset of ${1, 2, dots , 2019}$ such that no two members of $S$ differ by 1, 2 or 9. What is the...
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My solution is that you can take 4 integers from any 12 consecutive integers, and since $2019 = 168*12+3$ you'll have a total of $168*4+1 = 673$ integers in $S$. I have no way to check if my solution is correct so any kind of help will be appreciated.
combinatorics extremal-combinatorics
asked Jan 18 at 13:08
Shafin AhmedShafin Ahmed
707
$endgroup$
add a comment |
$begingroup$
My solution is that you can take 4 integers from any 12 consecutive integers, and since $2019 = 168*12+3$ you'll have a total of $168*4+1 = 673$ integers in $S$. I have no way to check if my solution is correct so any kind of help will be appreciated.
combinatorics extremal-combinatorics
asked Jan 18 at 13:08
Shafin AhmedShafin Ahmed
707
$endgroup$
$begingroup$
Problems with the current year appearing in them are common in math contests. Does this question come from some contest? Could you provide the source, please?
$endgroup$
– Arnaud D.
Jan 18 at 13:29
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@ArnaudD. This question is from a High School Math Olympiad hosted by the CSE dept. of Bangladesh University of Engineering and Technology
$endgroup$
– Shafin Ahmed
Jan 18 at 13:36
add a comment |
$begingroup$
My solution is that you can take 4 integers from any 12 consecutive integers, and since $2019 = 168*12+3$ you'll have a total of $168*4+1 = 673$ integers in $S$. I have no way to check if my solution is correct so any kind of help will be appreciated.
combinatorics extremal-combinatorics
asked Jan 18 at 13:08
Shafin AhmedShafin Ahmed
707
$endgroup$
My solution is that you can take 4 integers from any 12 consecutive integers, and since $2019 = 168*12+3$ you'll have a total of $168*4+1 = 673$ integers in $S$. I have no way to check if my solution is correct so any kind of help will be appreciated.
combinatorics extremal-combinatorics
combinatorics extremal-combinatorics
asked Jan 18 at 13:08
Shafin AhmedShafin Ahmed
707
asked Jan 18 at 13:08
Shafin AhmedShafin Ahmed
707
asked Jan 18 at 13:08
Shafin AhmedShafin Ahmed
707
asked Jan 18 at 13:08
Shafin AhmedShafin Ahmed
707
asked Jan 18 at 13:08
Shafin AhmedShafin Ahmed
707
707
$begingroup$
Problems with the current year appearing in them are common in math contests. Does this question come from some contest? Could you provide the source, please?
$endgroup$
– Arnaud D.
Jan 18 at 13:29
$begingroup$
@ArnaudD. This question is from a High School Math Olympiad hosted by the CSE dept. of Bangladesh University of Engineering and Technology
$endgroup$
– Shafin Ahmed
Jan 18 at 13:36
add a comment |
$begingroup$
Problems with the current year appearing in them are common in math contests. Does this question come from some contest? Could you provide the source, please?
$endgroup$
– Arnaud D.
Jan 18 at 13:29
$begingroup$
@ArnaudD. This question is from a High School Math Olympiad hosted by the CSE dept. of Bangladesh University of Engineering and Technology
$endgroup$
– Shafin Ahmed
Jan 18 at 13:36
$begingroup$
Problems with the current year appearing in them are common in math contests. Does this question come from some contest? Could you provide the source, please?
$endgroup$
– Arnaud D.
Jan 18 at 13:29
$begingroup$
Problems with the current year appearing in them are common in math contests. Does this question come from some contest? Could you provide the source, please?
$endgroup$
– Arnaud D.
Jan 18 at 13:29
$begingroup$
@ArnaudD. This question is from a High School Math Olympiad hosted by the CSE dept. of Bangladesh University of Engineering and Technology
$endgroup$
– Shafin Ahmed
Jan 18 at 13:36
$begingroup$
@ArnaudD. This question is from a High School Math Olympiad hosted by the CSE dept. of Bangladesh University of Engineering and Technology
$endgroup$
– Shafin Ahmed
Jan 18 at 13:36
add a comment |
2 Answers
2
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$begingroup$
Your answer has an issue where the numbers in one set of 12 consecutive numbers might not be compatible with the numbers from the next set of 12 consecutive numbers. For example, using the initial sequence of Farseed AF's answer, ${1, 4, 7, 11}$, you will no longer be able to choose $13$ from the $13$ to $24$. In fact, if you go further, between $25$ to $36$, you will not be able to find $4$ numbers.
To see another way that Fareed AF's answer is optimal, let's phrase the problem this way. Without loss of generality, let the set be ordered from smallest to largest and starting with 1. Then we can think of $S = {s_1,s_2,ldots,s_n}$ as a sequences of differences $D = {d_1,d_2,ldots,d_{n-1}}$ where $d_i = s_{i+1} - s_i$. And the restriction then becomes no consecutive sums of $d_i$ can be $1$, $2$ or $9$. To maximize the sequence length, we minimize the $d_i$. So we would want $d_1 = 3, d_2 = 3, d_3 = 4, d_4 = 3, ldots$ which indeed forms Fareed AF's sequence.
answered Jan 18 at 14:48
DubsDubs
52925
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add a comment |
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I might be wrong, but I think this set is the one with the most cardinal, because I think it has the least difference between its elements.
And this set is ${1,4,7,11,14,17,21,24,27,31,34,37....}$
we can notice that the last digit of each term will be either $1$,$4$, or $7$.
Which means from each 10 digits we will choose 3 digits.
So from 2019 digits we will choose 606 digits.
answered Jan 18 at 13:25
Fareed AFFareed AF
52612
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your answer has an issue where the numbers in one set of 12 consecutive numbers might not be compatible with the numbers from the next set of 12 consecutive numbers. For example, using the initial sequence of Farseed AF's answer, ${1, 4, 7, 11}$, you will no longer be able to choose $13$ from the $13$ to $24$. In fact, if you go further, between $25$ to $36$, you will not be able to find $4$ numbers.
To see another way that Fareed AF's answer is optimal, let's phrase the problem this way. Without loss of generality, let the set be ordered from smallest to largest and starting with 1. Then we can think of $S = {s_1,s_2,ldots,s_n}$ as a sequences of differences $D = {d_1,d_2,ldots,d_{n-1}}$ where $d_i = s_{i+1} - s_i$. And the restriction then becomes no consecutive sums of $d_i$ can be $1$, $2$ or $9$. To maximize the sequence length, we minimize the $d_i$. So we would want $d_1 = 3, d_2 = 3, d_3 = 4, d_4 = 3, ldots$ which indeed forms Fareed AF's sequence.
answered Jan 18 at 14:48
DubsDubs
52925
$endgroup$
add a comment |
$begingroup$
Your answer has an issue where the numbers in one set of 12 consecutive numbers might not be compatible with the numbers from the next set of 12 consecutive numbers. For example, using the initial sequence of Farseed AF's answer, ${1, 4, 7, 11}$, you will no longer be able to choose $13$ from the $13$ to $24$. In fact, if you go further, between $25$ to $36$, you will not be able to find $4$ numbers.
To see another way that Fareed AF's answer is optimal, let's phrase the problem this way. Without loss of generality, let the set be ordered from smallest to largest and starting with 1. Then we can think of $S = {s_1,s_2,ldots,s_n}$ as a sequences of differences $D = {d_1,d_2,ldots,d_{n-1}}$ where $d_i = s_{i+1} - s_i$. And the restriction then becomes no consecutive sums of $d_i$ can be $1$, $2$ or $9$. To maximize the sequence length, we minimize the $d_i$. So we would want $d_1 = 3, d_2 = 3, d_3 = 4, d_4 = 3, ldots$ which indeed forms Fareed AF's sequence.
answered Jan 18 at 14:48
DubsDubs
52925
$endgroup$
add a comment |
$begingroup$
Your answer has an issue where the numbers in one set of 12 consecutive numbers might not be compatible with the numbers from the next set of 12 consecutive numbers. For example, using the initial sequence of Farseed AF's answer, ${1, 4, 7, 11}$, you will no longer be able to choose $13$ from the $13$ to $24$. In fact, if you go further, between $25$ to $36$, you will not be able to find $4$ numbers.
To see another way that Fareed AF's answer is optimal, let's phrase the problem this way. Without loss of generality, let the set be ordered from smallest to largest and starting with 1. Then we can think of $S = {s_1,s_2,ldots,s_n}$ as a sequences of differences $D = {d_1,d_2,ldots,d_{n-1}}$ where $d_i = s_{i+1} - s_i$. And the restriction then becomes no consecutive sums of $d_i$ can be $1$, $2$ or $9$. To maximize the sequence length, we minimize the $d_i$. So we would want $d_1 = 3, d_2 = 3, d_3 = 4, d_4 = 3, ldots$ which indeed forms Fareed AF's sequence.
answered Jan 18 at 14:48
DubsDubs
52925
$endgroup$
Your answer has an issue where the numbers in one set of 12 consecutive numbers might not be compatible with the numbers from the next set of 12 consecutive numbers. For example, using the initial sequence of Farseed AF's answer, ${1, 4, 7, 11}$, you will no longer be able to choose $13$ from the $13$ to $24$. In fact, if you go further, between $25$ to $36$, you will not be able to find $4$ numbers.
To see another way that Fareed AF's answer is optimal, let's phrase the problem this way. Without loss of generality, let the set be ordered from smallest to largest and starting with 1. Then we can think of $S = {s_1,s_2,ldots,s_n}$ as a sequences of differences $D = {d_1,d_2,ldots,d_{n-1}}$ where $d_i = s_{i+1} - s_i$. And the restriction then becomes no consecutive sums of $d_i$ can be $1$, $2$ or $9$. To maximize the sequence length, we minimize the $d_i$. So we would want $d_1 = 3, d_2 = 3, d_3 = 4, d_4 = 3, ldots$ which indeed forms Fareed AF's sequence.
answered Jan 18 at 14:48
DubsDubs
52925
answered Jan 18 at 14:48
DubsDubs
52925
answered Jan 18 at 14:48
DubsDubs
52925
answered Jan 18 at 14:48
DubsDubs
52925
52925
add a comment |
add a comment |
$begingroup$
I might be wrong, but I think this set is the one with the most cardinal, because I think it has the least difference between its elements.
And this set is ${1,4,7,11,14,17,21,24,27,31,34,37....}$
we can notice that the last digit of each term will be either $1$,$4$, or $7$.
Which means from each 10 digits we will choose 3 digits.
So from 2019 digits we will choose 606 digits.
answered Jan 18 at 13:25
Fareed AFFareed AF
52612
$endgroup$
add a comment |
$begingroup$
I might be wrong, but I think this set is the one with the most cardinal, because I think it has the least difference between its elements.
And this set is ${1,4,7,11,14,17,21,24,27,31,34,37....}$
we can notice that the last digit of each term will be either $1$,$4$, or $7$.
Which means from each 10 digits we will choose 3 digits.
So from 2019 digits we will choose 606 digits.
answered Jan 18 at 13:25
Fareed AFFareed AF
52612
$endgroup$
add a comment |
$begingroup$
I might be wrong, but I think this set is the one with the most cardinal, because I think it has the least difference between its elements.
And this set is ${1,4,7,11,14,17,21,24,27,31,34,37....}$
we can notice that the last digit of each term will be either $1$,$4$, or $7$.
Which means from each 10 digits we will choose 3 digits.
So from 2019 digits we will choose 606 digits.
answered Jan 18 at 13:25
Fareed AFFareed AF
52612
$endgroup$
I might be wrong, but I think this set is the one with the most cardinal, because I think it has the least difference between its elements.
And this set is ${1,4,7,11,14,17,21,24,27,31,34,37....}$
we can notice that the last digit of each term will be either $1$,$4$, or $7$.
Which means from each 10 digits we will choose 3 digits.
So from 2019 digits we will choose 606 digits.
answered Jan 18 at 13:25
Fareed AFFareed AF
52612
edited Jan 20 at 14:04
answered Jan 18 at 13:25
Fareed AFFareed AF
52612
answered Jan 18 at 13:25
Fareed AFFareed AF
52612
answered Jan 18 at 13:25
Fareed AFFareed AF
52612
52612
add a comment |
add a comment |
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$begingroup$
Problems with the current year appearing in them are common in math contests. Does this question come from some contest? Could you provide the source, please?
$endgroup$
– Arnaud D.
Jan 18 at 13:29
$begingroup$
@ArnaudD. This question is from a High School Math Olympiad hosted by the CSE dept. of Bangladesh University of Engineering and Technology
$endgroup$
– Shafin Ahmed
Jan 18 at 13:36