A) 12
B) 21
C) 27
D) 30
Correct Answer: D
Solution :
[d]Given \[A=\{x:x\le 9,x\in N\}=\{1,2,3,4,5,6,7,8,9\}\] 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 cannot be express as such that \[a+b+c\]multiple of 3\[6\to 1+2+3\] |
\[9\to 2+3+4,5+3+1,6+2+1\] |
\[12\to 9+2+1,8+3+1,7+1+4,7+2+3\] |
\[6+4+2,\text{ }6+5+1,\text{ }5+4+3\] |
\[15\to 9+4+2,9+5+1,8+6+1,8+5+2,\] |
\[8+4+3,7+6+2,7+5+3,6+5+4\] |
\[18\to 9+8+1,9+7+2,9+6+3,\] |
\[9+5+4,8+7+3,8+6+4,7+6+5\] |
\[21\to 9+8+4,9+7+5,8+7+6\] |
\[24\to 9+8+7\] |
Hence, total largest possible subsets are 30. |
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