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Let $A=\{x \mid x \leq 9, x \in N\} .$ Let $B=\{a, b, c\}$ be the subset of $A$ where $(a+b+c)$ is a multiple of $3 .$ What is the largest possible number of subsets like B ?
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The correct answer is:
30
Given
$\mathrm{A}=\{\mathrm{x}: \mathrm{x} \leq 9, \mathrm{x} \in \mathrm{N}\}=\{1,2,3,4,5,6,7,8,9\}$
Total possible multiple of 3 are
$3,6,9,12,15,18,21,24,27$
But 3 and 27 are not possible because 3 and 27 can not be express as such that $a+b+c$ is multiple of 3 $6 \rightarrow 1+2+3$
$9 \rightarrow 2+3+4,5+3+1,6+2+1$
$12 \rightarrow 9+2+1,8+3+1,7+1+4,7+2+3$
$6+4+2,6+5+1,5+4+3$
$15 \rightarrow 9+4+2,9+5+1,8+6+1,8+5+2$,
$8+4+3,7+6+2,7+5+3,6+5+4$
$18 \rightarrow 9+8+1,9+7+2,9+6+3$
$9+5+4,8+7+3,8+6+4,7+6+5$
$21 \rightarrow 9+8+4,9+7+5,8+7+6$
$24 \rightarrow 9+8+7$
Hence, total largest possible subsets are 30 .
$\mathrm{A}=\{\mathrm{x}: \mathrm{x} \leq 9, \mathrm{x} \in \mathrm{N}\}=\{1,2,3,4,5,6,7,8,9\}$
Total possible multiple of 3 are
$3,6,9,12,15,18,21,24,27$
But 3 and 27 are not possible because 3 and 27 can not be express as such that $a+b+c$ is multiple of 3 $6 \rightarrow 1+2+3$
$9 \rightarrow 2+3+4,5+3+1,6+2+1$
$12 \rightarrow 9+2+1,8+3+1,7+1+4,7+2+3$
$6+4+2,6+5+1,5+4+3$
$15 \rightarrow 9+4+2,9+5+1,8+6+1,8+5+2$,
$8+4+3,7+6+2,7+5+3,6+5+4$
$18 \rightarrow 9+8+1,9+7+2,9+6+3$
$9+5+4,8+7+3,8+6+4,7+6+5$
$21 \rightarrow 9+8+4,9+7+5,8+7+6$
$24 \rightarrow 9+8+7$
Hence, total largest possible subsets are 30 .
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