a) Suppose that $n$ is an odd integer. Prove that $k(n-k)$ is divisible by $2$ for all positive integers $k$.

b) Find an integer $k$ such that $k(100-k)$ is not divisible by $11$.

c) Suppose that $p$ is an odd prime, and $n$ is an integer. 
Prove that there is an integer $k$ such that $k(n-k)$ is not divisible by $p$. 

d) Suppose that $p,q$ are two different odd primes, and $n$ is an integer. 
Prove that there is an integer $k$ such that $k(n-k)$ is not divisible by any of $p,q$.
### Part (a)
1. Let \( n \) be an odd integer. We need to prove that \( k(n-k) \) is divisible by 2 for all positive integers \( k \).
2. Consider the parity (odd or even nature) of \( k \):
   - If \( k \) is even, then \( k \) is divisible by 2, and hence \( k(n-k) \) is divisible by 2.
   - If \( k \) is odd, then \( n-k \) is even because \( n \) is odd and the difference between an odd number and an odd number is even. Therefore, \( n-k \) is divisible by 2, and hence \( k(n-k) \) is divisible by 2.
3. In both cases, \( k(n-k) \) is divisible by 2.

\(\blacksquare\)

### Part (b)
1. We need to find an integer \( k \) such that \( k(100-k) \) is not divisible by 11.
2. Consider the expression \( k(100-k) \mod 11 \):
   - We need \( k(100-k) \not\equiv 0 \mod 11 \).
3. Let's test \( k = 2 \):
   - \( k(100-k) = 2(100-2) = 2 \times 98 = 196 \).
   - \( 196 \mod 11 = 9 \).
4. Since \( 9 \not\equiv 0 \mod 11 \), \( k = 2 \) is a valid solution.

\(\boxed{k = 2}\)

### Part (c)
1. Let \( p \) be an odd prime and \( n \) be an integer. We need to prove that there is an integer \( k \) such that \( k(n-k) \) is not divisible by \( p \).
2. Consider the residues of \( k \) and \( n \) modulo \( p \):
   - Let \( k \equiv a \mod p \) and \( n \equiv b \mod p \).
3. Since \( p \) is an odd prime, the residues modulo \( p \) are \( 0, 1, 2, \ldots, p-1 \).
4. We need to choose \( k \) such that \( k(n-k) \not\equiv 0 \mod p \):
   - Avoid \( k \equiv 0 \mod p \) because \( k \equiv 0 \mod p \) implies \( k(n-k) \equiv 0 \mod p \).
   - Choose \( k \) such that \( a \neq b \mod p \).
5. Since there are \( p \) residues and only one residue \( b \) to avoid, there are \( p-1 \) choices for \( k \) such that \( k(n-k) \not\equiv 0 \mod p \).

\(\blacksquare\)

### Part (d)
1. Let \( p \) and \( q \) be two different odd primes, and \( n \) be an integer. We need to prove that there is an integer \( k \) such that \( k(n-k) \) is not divisible by any of \( p \) or \( q \).
2. Consider the residues of \( k \) and \( n \) modulo \( p \) and \( q \):
   - Let \( k \equiv a \mod p \) and \( n \equiv b \mod p \).
   - Let \( k \equiv c \mod q \) and \( n \equiv d \mod q \).
3. We need to choose \( k \) such that \( k(n-k) \not\equiv 0 \mod p \) and \( k(n-k) \not\equiv 0 \mod q \):
   - Avoid \( k \equiv 0 \mod p \) and \( k \equiv b \mod p \).
   - Avoid \( k \equiv 0 \mod q \) and \( k \equiv d \mod q \).
4. Since \( p \) and \( q \) are different primes, the Chinese Remainder Theorem guarantees that there are solutions to the system of congruences:
   - \( k \not\equiv 0 \mod p \)
   - \( k \not\equiv b \mod p \)
   - \( k \not\equiv 0 \mod q \)
   - \( k \not\equiv d \mod q \)
5. Therefore, there exists an integer \( k \) such that \( k(n-k) \) is not divisible by either \( p \) or \( q \).

\(\blacksquare\)