1. The problem involves understanding the behavior of the function $A(n)$, which counts the number of prime numbers from 1 up to $n$. 2. We analyze each statement: - Statement 1: "$A(k)$ is sometimes greater than $k$." Since $A(k)$ counts primes up to $k$, it cannot be greater than $k$ because the number of primes up to $k$ is at most $k$. So this statement is false. - Statement 2: "There is always a counting number $m$ such that $A(k)$ does not divide $21m + 3$ for any $k$." This means for some $m$, no $A(k)$ divides $21m + 3$. We will check this later. - Statement 3: "$A(k)$ is always an odd number." The number of primes up to $k$ varies and can be even or odd. For example, $A(2) = 1$ (odd), $A(3) = 2$ (even). So this statement is false. - Statement 4: "For any counting number $m$, there is a counting number $k$ such that $A(k)$ divides $21m + 3$." This means for every $m$, we can find $k$ with $A(k) ig| 21m + 3$. 3. To analyze statements 2 and 4, consider the values of $A(k)$ and divisibility: - $A(k)$ takes values in the sequence of prime counts: 0,1,2,3,5,7,11,13,... - Since $A(k)$ grows without bound, for any fixed $m$, $21m + 3$ is fixed. - Statement 4 claims for any $m$, some $A(k)$ divides $21m + 3$. - Statement 2 claims there exists $m$ such that no $A(k)$ divides $21m + 3$. 4. Since $A(k)$ can be arbitrarily large, and $21m + 3$ is fixed for each $m$, it is possible that no $A(k)$ divides $21m + 3$ for some $m$ if $21m + 3$ is small or has prime factors not in the range of $A(k)$. 5. However, $A(k)$ includes 1 (since $A(2)=1$), and 1 divides every integer, so for any $m$, $A(2)=1$ divides $21m + 3$. 6. Therefore, statement 4 is true because $A(2)=1$ divides all $21m + 3$. 7. Statement 2 is false because there is no $m$ such that no $A(k)$ divides $21m + 3$ (since $A(2)=1$ divides all). Final answers: - Statement 1: False - Statement 2: False - Statement 3: False - Statement 4: True