Q33 / 58 · KVPY 2016 (SA) · Mathematics
Let $A_{1}, A_{2} \ldots, A_{m}$ be non-empty subsets of $\{1,2,3, \ldots \ldots, 100\}$ satisfying the following conditions:
(1) the numbers $\left|\mathrm{A}_{1}\right|,\left|\mathrm{A}_{2}\right|, \ldots,\left|\mathrm{A}_{\mathrm{m}}\right|$ are distinct;
(2) $\mathrm{A}_{1}, \mathrm{~A}_{2}, \ldots, \mathrm{A}_{\mathrm{m}}$ are pairwise disjoint.
(Here $\mid \mathrm{A}$ denotes the number of elements in the set $\mathrm{A}$.) Then the maximum possible value of $\mathrm{m}$ is
- Correct answerA13
- B14
- C15
- D16
Solution
The possibility is
$\begin{array}{l}
\left|\mathrm{A}_{1}\right|=1 ;\left|\mathrm{A}_{2}\right|=2 ;\left|\mathrm{A}_{3}\right|=3 ; \ldots \ldots .,\left|\mathrm{A}_{m}\right|=\mathrm{m} \\
1+2+3+\ldots \ldots+\mathrm{m} \leq 100 \quad\{\text { Because all are disjoint }\} \\
\Rightarrow \frac{\mathrm{m}(\mathrm{m}+1)}{2} \leq 100 \\
\Rightarrow \mathrm{m} < 14
\end{array}$
$14^{\text {th }}$ set will have the same size as that of one of the previous sets
So, $\mathrm{m}=13$
Hence, the correct answer is option: 'A'.