JEE Mathematics 2023 Question with Solution

Given: $S = \{1, 2, 3, 5, 7, 10, 11\}$

Find: The number of non-empty subsets of $S$ whose element-sum is divisible by $3$.

Classify the elements according to their remainders modulo $3$:

$$3k = 3$$ $$3k+1 = 1, 7, 10$$ $$3k+2 = 2, 5, 11$$

Now count valid subsets by size as shown in the solution.

Subsets containing one element:

$$S^1 = 1$$

Subsets containing two elements:

$${}^3C_1 \times {}^3C_1 = 9$$

Subsets containing three elements:

$${}^3C_1 \times {}^3C_1 + 1 + 1 = 11$$

Subsets containing four elements:

$${}^3C_3 + {}^3C_3 + {}^3C_2 \times {}^3C_2 = 11$$

Subsets containing five elements:

$${}^3C_2 \times {}^3C_2 \times 1 = 9$$

Subsets containing six elements:

$$S^6 = 1$$

Subsets containing seven elements:

$$S^7 = 1$$

Therefore,

$$\text{sum} = 43$$

So, the number of non-empty subsets is $43$.

Given: $S = \{1, 2, 3, 5, 7, 10, 11\}$

Find: The number of non-empty subsets whose sum is divisible by $3$.

First compute residues modulo $3$:

$$1 \equiv 1,\quad 2 \equiv 2,\quad 3 \equiv 0,\quad 5 \equiv 2,\quad 7 \equiv 1,\quad 10 \equiv 1,\quad 11 \equiv 2 \pmod 3$$

Hence the groups are:

$$\text{mod } 3 = 0 : \{3\},\quad \text{mod } 3 = 1 : \{1,7,10\},\quad \text{mod } 3 = 2 : \{2,5,11\}$$

A subset sum is divisible by $3$ when the total residue is $0 \pmod 3$. The extracted solution counts such subsets size-wise and gives:

• size $1$: $1$ subset
• size $2$: $9$ subsets
• size $3$: $11$ subsets
• size $4$: $11$ subsets
• size $5$: $9$ subsets
• size $6$: $1$ subset
• size $7$: $1$ subset

Adding these counts,

$$1 + 9 + 11 + 11 + 9 + 1 + 1 = 43$$

Therefore, the required number of non-empty subsets is $43$.