7. Peggy picks three positive integers between $1$ and $25$, inclusive, and tells us the following information about those numbers:
[list]
[*] Exactly one of them is a multiple of $2$;
[*] Exactly one of them is a multiple of $3$;
[*] Exactly one of them is a multiple of $5$;
[*] Exactly one of them is a multiple of $7$;
[*] Exactly one of them is a multiple of $11$.
[/list]
What is the maximum possible sum of the integers that Peggy picked?

8. What is the largest positive integer $k$ such that $2^k$ divides $2^{4^8}+8^{2^4}+4^{8^2}$?

9. Find the smallest integer $n$ such that $n$ is the sum of $7$ consecutive positive integers and the sum of $12$ consecutive positive integers.
### Problem 7:
Peggy picks three positive integers between $1$ and $25$, inclusive, with the following conditions:
- Exactly one of them is a multiple of $2$.
- Exactly one of them is a multiple of $3$.
- Exactly one of them is a multiple of $5$.
- Exactly one of them is a multiple of $7$.
- Exactly one of them is a multiple of $11$.

We need to find the maximum possible sum of these integers.

1. **Identify the numbers that satisfy multiple conditions:**
   - The number $22$ is a multiple of $2$ and $11$.
   - The number $21$ is a multiple of $3$ and $7$.
   - The number $25$ is a multiple of $5$.

2. **Verify that these numbers meet the conditions:**
   - $22$ is a multiple of $2$ and $11$.
   - $21$ is a multiple of $3$ and $7$.
   - $25$ is a multiple of $5$.

3. **Sum these numbers:**
   \[
   22 + 21 + 25 = 68
   \]

Thus, the maximum possible sum of the integers that Peggy picked is $\boxed{68}$.

### Problem 8:
We need to find the largest positive integer $k$ such that $2^k$ divides $2^{4^8} + 8^{2^4} + 4^{8^2}$.

1. **Simplify the expression:**
   \[
   2^{4^8} + 8^{2^4} + 4^{8^2}
   \]
   - $2^{4^8}$ is already in the form of $2$ raised to a power.
   - $8^{2^4} = (2^3)^{16} = 2^{48}$.
   - $4^{8^2} = (2^2)^{64} = 2^{128}$.

2. **Rewrite the expression:**
   \[
   2^{4^8} + 2^{48} + 2^{128}
   \]

3. **Factor out the smallest power of $2$:**
   \[
   2^{48}(2^{4^8 - 48} + 1 + 2^{128 - 48})
   \]
   - $4^8 = 65536$, so $2^{4^8 - 48} = 2^{65536 - 48} = 2^{65488}$.
   - $2^{128 - 48} = 2^{80}$.

4. **The expression becomes:**
   \[
   2^{48}(2^{65488} + 1 + 2^{80})
   \]

Since $2^{48}$ is factored out, the largest $k$ such that $2^k$ divides the expression is $48$.

Thus, the answer is $\boxed{48}$.

### Problem 9:
Find the smallest integer $n$ such that $n$ is the sum of $7$ consecutive positive integers and the sum of $12$ consecutive positive integers.

1. **Sum of $7$ consecutive integers:**
   Let the integers be $a, a+1, a+2, \ldots, a+6$. The sum is:
   \[
   7a + 21
   \]
   Therefore, $n \equiv 0 \pmod{7}$.

2. **Sum of $12$ consecutive integers:**
   Let the integers be $b, b+1, b+2, \ldots, b+11$. The sum is:
   \[
   12b + 66
   \]
   Therefore, $n \equiv 6 \pmod{12}$.

3. **Solve the system of congruences:**
   \[
   n \equiv 0 \pmod{7}
   \]
   \[
   n \equiv 6 \pmod{12}
   \]

4. **Find the smallest $n$ that satisfies both conditions:**
   - $n = 7k$ for some integer $k$.
   - $7k \equiv 6 \pmod{12}$.

   Testing values of $k$:
   - For $k = 6$, $7 \times 6 = 42 \equiv 6 \pmod{12}$.

   So, $n = 42$ satisfies both conditions. However, $n$ must be greater than $66$ (since $12b + 66$ must be positive).

5. **Find the next solution:**
   - $n = 42 + 84 = 126$ (since $84$ is the least common multiple of $7$ and $12$).

Thus, the smallest integer $n$ is $\boxed{126}$.