Question

We have given a list of 2024 positive integers, that has a unique mode, which occurs exactly 11 times. Let $N$ is the least number of distinct values that can occur in the list. Then the value of $\frac{N}{7}$ is

Answer 1

29

Answer 2

Let $S$ be the total number of integers in the list, so $S = 2024$. Let $m$ be the unique mode of the list, and its frequency is given as $f_m = 11$. Since $m$ is the unique mode, its frequency must be strictly greater than the frequency of any other distinct value in the list. Let the distinct values in the list be $v_1, v_2, \dots, v_k$, with frequencies $f_1, f_2, \dots, f_k$ respectively. Let $v_1 = m$, so $f_1 = 11$. For any other distinct value $v_i$ where $i \neq 1$, its frequency $f_i$ must satisfy $f_i < f_1$. Thus, $f_i \le 10$ for all $i = 2, 3, \dots, k$. Also, for a value to be present in the list as a distinct value, its frequency must be at least 1. So, $f_i \ge 1$ for all $i = 2, 3, \dots, k$.

The sum of the frequencies of all distinct values is equal to the total number of integers in the list: $\sum_{i=1}^k f_i = S$

$f_1 + \sum_{i=2}^k f_i = 2024$

$11 + \sum_{i=2}^k f_i = 2024$

$\sum_{i=2}^k f_i = 2024 - 11 = 2013$.

We want to find the least number of distinct values, $N$, which is the minimum possible value of $k$. The number of distinct values other than the mode is $k-1$. Let this be $M = k-1$. The sum of the frequencies of these $M$ distinct values is 2013: $\sum_{i=2}^{k} f_i = f_2 + f_3 + \dots + f_{M+1} = 2013$. For each of these frequencies, we know that $1 \le f_i \le 10$. To minimize the number of distinct values other than the mode, $M$, we need to maximize the individual frequencies $f_i$. The maximum possible value for each $f_i$ is 10. The sum of $M$ frequencies, each at most 10, is at most $M \times 10$. So, $2013 = \sum_{i=2}^{M+1} f_i \le M \times 10$.

$2013 \le 10M$

$M \ge \frac{2013}{10} = 201.3$. Since $M$ must be an integer (the number of distinct values), the minimum possible value for $M$ is 202. So, the minimum number of distinct values other than the mode is 202. The total number of distinct values is $k = 1 + M$. The least number of distinct values is $N = k_{min} = 1 + M_{min} = 1 + 202 = 203$.

We need to verify that it is possible to construct a list with 203 distinct values satisfying the conditions. If $k=203$, then $M=k-1=202$. We need to find 202 frequencies $f_2, f_3, \dots, f_{203}$ such that $1 \le f_i \le 10$ and their sum is 2013. We can express 2013 in terms of multiples of 10: $2013 = 10 \times 201 + 3$. This suggests we can have 201 distinct values with frequency 10 and one distinct value with frequency 3. Let $f_2 = f_3 = \dots = f_{202} = 10$ (201 values) and $f_{203} = 3$ (1 value). The sum of these frequencies is $201 \times 10 + 3 = 2010 + 3 = 2013$. All these frequencies (10 and 3) are between 1 and 10, so they are valid frequencies for distinct values other than the mode. The distinct values could be, for example: The mode $v_1 = 1$ with frequency $f_1 = 11$. Other distinct values $v_2, \dots, v_{202}$ are $2, \dots, 202$ with frequencies $f_2 = \dots = f_{202} = 10$. The distinct value $v_{203} = 203$ with frequency $f_{203} = 3$. The total number of integers in the list is $11 + (201 \times 10) + 3 = 11 + 2010 + 3 = 2024$. The distinct values are $1, 2, \dots, 203$. There are $1 + 202 = 203$ distinct values. The frequencies are 11, 10 (201 times), and 3. The highest frequency is 11, and it corresponds to the value 1, which is the unique mode. This construction satisfies all the given conditions. So, the least number of distinct values is $N = 203$.

The question asks for the value of $\frac{N}{7}$. $\frac{N}{7} = \frac{203}{7}$. $203 \div 7 = 29$.

The final answer is $\frac{203}{7} = 29$.