Given a prime number [tex]\( p \)[/tex] and a positive integer [tex]\( k \)[/tex], we know that [tex]\( p \)[/tex] divides [tex]\( k^2 \)[/tex]. We need to determine which of the following is definitely divisible by [tex]\( p \)[/tex]:
[tex]\[
\begin{array}{|l|l|l|l|}
\hline
\frac{k}{2} & k & 7k & k^3 \\
\hline
\end{array}
\][/tex]
### Step-by-Step Solution
1. Given Condition Analysis:
- We are given that [tex]\( p \)[/tex] divides [tex]\( k^2 \)[/tex]. This means [tex]\( p \mid k^2 \)[/tex].
- A prime number [tex]\( p \)[/tex] that divides a square of a number implies [tex]\( p \)[/tex] must also divide the number itself (since a prime factor of a square must appear an even number of times in its prime factorization, and thus it must be present in the factorization of the original number as well).
Formally, if [tex]\( p \mid k^2 \)[/tex], then [tex]\( p \mid k \)[/tex].
2. Checking Divisibility:
- For [tex]\(\frac{k}{2}\)[/tex]:
[tex]\[
\frac{k}{2} \ \text{is not necessarily an integer}. \text{ Even if it is an integer, there is no guarantee that } p \text{ divides } \frac{k}{2}.
\][/tex]
Therefore, [tex]\(\frac{k}{2}\)[/tex] is not definitely divisible by [tex]\( p \)[/tex].
- For [tex]\( k \)[/tex]:
[tex]\[
\text{Since } p \mid k \ (\text{from our previous analysis}), \ k \ \text{is definitely divisible by } p.
\][/tex]
- For [tex]\( 7k \)[/tex]:
[tex]\[
\text{Multiplying } k \ \text{by a constant } 7 \ \text{does not affect the divisibility by } p.
\][/tex]
Thus, since [tex]\( p \mid k \)[/tex], [tex]\( p \)[/tex] also divides [tex]\( 7k \)[/tex].
- For [tex]\( k^3 \)[/tex]:
[tex]\[
k^3 = k \cdot k^2. \text{ Since } p \mid k \ \text{and } p \mid k^2, \ p \text{ must also divide } k^3 \ (\text{because } p \text{ divides each factor}).
\][/tex]
3. Conclusion:
Although [tex]\( k \)[/tex], [tex]\( 7k \)[/tex], and [tex]\( k^3 \)[/tex] are all divisible by [tex]\( p \)[/tex], the simplest and most direct answer is [tex]\( k \)[/tex].
Therefore, the number that is _definitely_ divisible by [tex]\( p \)[/tex] is:
[tex]\[
\boxed{k}
\][/tex]
