Set- i do not understand the question

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The discussion centers on finding the maximum size of a subset, P, from the set {1, 2, 3, ..., 50}, ensuring that no two distinct elements in P have a sum divisible by 7. Participants analyze which numbers can coexist in P based on their remainders when divided by 7, specifically focusing on avoiding pairs that sum to multiples of 7. They identify that if certain numbers are included in P, others must be excluded to maintain the condition. The conversation highlights the importance of modular arithmetic in determining valid combinations. Ultimately, the goal is to maximize the size of subset P while adhering to these constraints.
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Homework Statement



What is the maximum size of a subset, P, of {1, 2, 3, . . . , 50}
with the property that no pair of distinct elements of P
has a sum divisible by 7?

Homework Equations


The Attempt at a Solution



smallest sum = 3
largest sum = 99

factors of 7:

7 14 21 28 35 42 49 56 63 70 77 84 91 98

7 can be form by:

1 6
2 5
3 4
 
Last edited:
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Hi icystrike! :smile:
icystrike said:
What is the maximum size of a subset, P, of {1, 2, 3, . . . , 50}
with the property that no pair of distinct elements of P
has a sum divisible by 7?

7 can be form by:

1 6
2 5
3 4

ok … so if 1 is in P, what can't be in P? :wink:
 
1 2 3 4 5 7 8 9 10 11 12 14 15 16 17 18 19 21 22 23 24 25 26 28 29 30 31 32 33 35 36 37 38 39 40 42 43 44 45 46 47 49 50 !
yes
 
Last edited:
icystrike said:
1 2 3 4 5 7 8 9 10 11 12 14 15 16 17 18 19 21 22 23 24 25 26 28 29 30 31 32 33 35 36 37 38 39 40 42 43 44 45 46 47 49 50 !
yes

no, that's what could be in P … what can't be in P are 6 13 20 27 34 41 48, in other words everything = -1 (mod 7)

suppose 2 and 3 are also in P (as well as 1) … now what can't be in P? :smile:
 
tiny-tim said:
no, that's what could be in p … what can't be in p are 6 13 20 27 34 41 48, in other words everything = -1 (mod 7)

suppose 2 and 3 are also in p (as well as 1) … now what can't be in p? :smile:

5 12 19 26 33 40 47
4 11 18 25 32 39 46
 
icystrike said:
5 12 19 26 33 40 47
4 11 18 25 32 39 46

ok, so if 1 2 and 3 are in, then all the mod6s 5s and 4s are out …

can the other mod 1s 2s and 3s be in?
 
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