Let S = {1, 2, 3, ldots, 40} and let A be a s | vedclass

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(C) To find the maximum number of elements in $A$,we partition the set $S$ into subsets based on their remainders when divided by $5$:$R_0 = \{5, 10,

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$17$

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(C) To find the maximum number of elements in $A$,we partition the set $S$ into subsets based on their remainders when divided by $5$:$R_0 = \{5, 10, 15, 20, 25, 30, 35, 40\}$ (remainder $0$,size $8$)$R_1 = \{1, 6, 11, 16, 21, 26, 31, 36\}$ (remainder $1$,size $8$)$R_2 = \{2, 7, 12, 17, 22, 27, 32, 37\}$ (remainder $2$,size $8$)$R_3 = \{3, 8, 13, 18, 23, 28, 33, 38\}$ (remainder $3$,size $8$)$R_4 = \{4, 9, 14, 19, 24, 29, 34, 39\}$ (remainder $4$,size $8$)For the sum of any two elements in $A$ not to be divisible by $5$,we must select elements such that their remainders do not sum to $5$ or $0$ (mod $5$).$1$. We can pick at most one element from $R_0$ (if we pick two,their sum is divisible by $5$).$2$. We can pick all elements from $R_1$ (since $1+1=2, 1+6=7$,etc.,none are divisible by $5$ except $1+4$,but we don't pick from $R_4$).$3$. We can pick all elements from $R_2$ (since $2+2=4$,none are divisible by $5$ except $2+3$,but we don't pick from $R_3$).Thus,we choose all elements from $R_1$ ($8$ elements),all elements from $R_2$ ($8$ elements),and one element from $R_0$ ($1$ element).Total elements = $8 + 8 + 1 = 17$.

Let $S = \{1, 2, 3, \ldots, 40\}$ and let $A$ be a subset of $S$ such that no two elements in $A$ have their sum divisible by $5$. What is the maximum number of elements possible in $A$?

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