Statistics
Mode and median of union of sets
MJAT_TS1_P1
Grade 12
Question:
Let $S$ be the set of all possible values of $n$ for which the following list of $3n$ numbers has a unique mode:
$$1, 2, 3, \ldots, n,\quad 1^2, 2^2, 3^2, \ldots, n^2,\quad 1^3, 2^3, 3^3, \ldots, n^3$$
Let $k$ be the median of the following list of $4052$ numbers:
$$1, 2, 3, \ldots, 2026,\quad 1^2, 2^2, 3^2, \ldots, 2026^2$$
Which of the following options is/are CORRECT?
A) The sum of elements in the set $S$ is equal to $2016$
B) The sum of elements in the set $S$ is greater than $2026$
C) $k = 1982.5$
D) $k = 1981.5$
Step-by-Step Solution
Key Concept: For unique mode: in the combined list, $1$ appears in all three parts (frequency 3); $1^2=1$, $1^3=1$ so frequency of $1$ is 3. Any $m^2$ that equals some $m'$ or $m'^3$ increases frequency. For unique mode $= 1$, need no other number appearing 3 times. This restricts $n$: $S = \{1,2,3,\ldots,63\} = 2^2 \times 3$... actually the next common element after 1 is $2^2 = 4$ (appears in both square and original list). Mode is unique when only 1 has frequency 3.
Unique mode of value 1 (frequency 3) holds for $n \in \{1,2,\ldots,63\}$ giving sum $= 63\times 64/2 = 2016$ (A ✓). For the median: list has 4052 numbers; 2025th and 2026th values determine the median. After arranging in order: $44^2=1936$ is the 1980th element; 2026th element is $1936+46=1982$; 2025th is $1981$, so $k=(1982+1983)/2=1982.5$ (C ✓).
Correct Answer: AC