Set- i do not understand the question

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Homework Help Overview

The problem involves finding the maximum size of a subset, P, from the set {1, 2, 3, ..., 50} such that no pair of distinct elements in P has a sum that is divisible by 7. The discussion revolves around understanding the implications of including certain numbers in the subset and the resulting exclusions based on modular arithmetic.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of including specific numbers in the subset and what numbers must be excluded as a result. There are attempts to identify pairs of numbers that sum to multiples of 7 and to explore the modular relationships among the numbers.

Discussion Status

The discussion is active, with participants questioning the inclusion of certain numbers and exploring the consequences of those choices. There is a focus on identifying which numbers cannot be included in the subset based on the modular conditions established.

Contextual Notes

Participants reference modular arithmetic, specifically modulo 7, to determine which numbers can coexist in the subset without violating the problem's conditions. There is an ongoing exploration of the relationships between the numbers and their residues modulo 7.

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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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