Question

Let $A=\{x|x\le 9,x\in N\}$. 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 $B$?
A. $12$
B. $21$
C. $27$
D. $30$

Answer 1

Hint: To solve this question we will first determine all the multiples of $3$ which can be expressed as $(a+b+c)$ with the elements of set $A=\{x|x\le 9,x\in N\}$. We will then determine all the possible subsets of each of the multiple and then add them to find the largest possible number of subsets like $B$.

Complete step by step solution: We have given a set $A$ such that $A=\{x|x\le 9,x\in N\}$ an a subset $B=\{a,b,c\}$ where $(a+b+c)$ is a multiple of $3$ and we have to find the largest possible number of subsets like $B$.
The set $A=\{x|x\le 9,x\in N\}$ means $A=\{1,2,3,4,5,6,7,8,9\}$ . We will first find all the multiples of $3$ which can be expressed as $(a+b+c)$.
The multiples are: $3,6,9,12,15,18,21,24,27,30,...$. But we will only include the multiples $6,9,12,15,18,21,24$because rest cannot be expressed as $(a+b+c)$. We will now form the largest possible number of subsets from set $A=\{1,2,3,4,5,6,7,8,9\}$ which will be multiple of $3$and expressed as $(a+b+c)$.
$\begin{align}
  & 6\to 1+2+3 \\
 & 9\to 2+3+4\,\,\,,5+3+1,\,\,6+2+1 \\
 & 12\to 9+2+1,\,\,8+3+1,\,\,7+4+1,\,\,6+5+1,\,\,7+3+2,\,\,\,6+4+2,\,\,5+4+3 \\
 & 15\to 9+4+2,\,\,9+5+1,\,\,8+6+1,\,\,8+5+2,\,\,8+4+3,\,\,7+6+2,\,\,7+5+3,\,\,6+5+4 \\
 & 18\to 9+8+1,\,\,9+7+2,\,\,9+6+3,\,\,9+5+4,\,\,8+7+3,\,\,8+6+4,\,\,7+6+5 \\
 & 21\to 9+8+4,\,\,9+7+5,\,\,8+7+6 \\
 & 24\to 9+8+7 \\
\end{align}$
The number of subset of multiple $6$ is $1$, $9$ is $3$, $12$ is $7$, $15$ is $8$, $18$ is $7$, $21$ is $3$ and $24$ is $1$.
Now we will now add all number of subsets of each of the multiple to find the largest possible number of subsets.
$\begin{align}
  & =1+3+7+8+7+3+1 \\
 & =30 \\
\end{align}$
The largest possible number of subsets like $B=\{a,b,c\}$which is a subset of $A=\{x|x\le 9,x\in N\}$ can be expressed as $(a+b+c)$ and is a multiple of $3$ is $30$.

Option ‘D’ is correct

Note: