JEE Main 2024 — 27 January, Evening Shift. Previous Year Question.
Problem
Let $A$ and $B$ be two finite sets with $m$ and $n$ elements respectively. The total number of subsets of the set $A$ is $56$ more than the total number of subsets of $B$. Then the distance of the point $P(m,n)$ from the point $Q(-2,-3)$ is:
Key insight. $2^m-2^n=56$ isn’t an equation to solve by trial and error forever — factoring out $2^n$ leaves $2^n(2^{m-n}-1)=56$, where the second factor is always odd. Since $56=2^3\times7$, this immediately forces $2^n=8$ and $2^{m-n}-1=7$, pinning down both $m$ and $n$ uniquely.
Watch this explained step by step → · More on the Shiwam’s Classes channel
Approach
Write the number of subsets of each set as $2^m$ and $2^n$, and translate the given condition into $2^m-2^n=56$. Factor out $2^n$ to isolate an odd factor, match it against the prime factorisation of $56$, and solve for $m$ and $n$. Use these as the coordinates of $P$ and apply the distance formula to find $PQ$.
Solution
Step 1 — Set up the equation for subset counts
$$2^m – 2^n = 56$$
Step 2 — Factor out 2ⁿ
$$2^n(2^{m-n}-1) = 56$$
Step 3 — Use the fact that 2^(m−n) − 1 is odd
Since $2^{m-n}-1$ is always odd (one less than a power of $2$), and $56=2^3\times7$, the entire power of $2$ in $56$ must come from $2^n$:
$$2^n = 8 \implies n=3$$
Step 4 — Solve for m
$$2^{m-n}-1 = 7 \implies 2^{m-n}=8 \implies m-n=3 \implies m=6$$
Step 5 — Identify point P
$$P = (m,n) = (6,3)$$
Step 6 — Apply the distance formula
$$PQ = \sqrt{(6-(-2))^2+(3-(-3))^2} = \sqrt{8^2+6^2} = \sqrt{64+36} = \sqrt{100} = 10$$
Answer
$$10$$
Common mistakes
- Trying to solve $2^m-2^n=56$ by guessing values of $m$ and $n$ directly without factoring. This can work by luck, but factoring out $2^n$ and using the odd/even structure of $56=2^3\times7$ guarantees finding the unique solution systematically.
- Mixing up which coordinate is $m$ and which is $n$ when substituting into the distance formula. Since $P=(m,n)=(6,3)$ specifically (not $(3,6)$), swapping them would still give the same distance here by coincidence of the numbers involved, but it’s worth being careful about which set’s subset-count difference determines which coordinate.
Practise next
- Two finite sets $C$ and $D$ have $p$ and $q$ elements, with the number of subsets of $C$ being $24$ more than that of $D$. Find the distance from $(p,q)$ to $(1,-2)$, using the same factoring approach.
Show answer
$\sqrt{41}$. The subset counts are $2^p$ and $2^q$, so $2^p-2^q=24$, that is $2^q\left(2^{p-q}-1\right)=24=8\times3$.
The first factor is a power of $2$ and the second is odd, so they must be $8$ and $3$: $q=3$ and $2^{p-q}=4$, giving $p=5$.
The distance from $(5,3)$ to $(1,-2)$ is $\sqrt{4^2+5^2}=\sqrt{41}$. Splitting $24$ into its power-of-two part and its odd part is what makes the solution unique.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.