Two natural numbers, $p$ and $q$, do not end in zero. The product of any pair, p and q, is a power of 10
(that is, $10, 100, 1000, 10 000$ , ...). If $p  >q$, the last digit of $p – q$ cannot be

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 9$
1. Given two natural numbers \( p \) and \( q \) that do not end in zero, and their product \( p \cdot q \) is a power of 10, i.e., \( 10^k \) for some integer \( k \).
2. Since \( p \cdot q = 10^k \), we can express \( p \) and \( q \) in terms of their prime factors. Specifically, \( 10^k = 2^k \cdot 5^k \). Therefore, one of the numbers must be a power of 2 and the other must be a power of 5.
3. Let \( p = 5^a \) and \( q = 2^b \) where \( a \) and \( b \) are positive integers. Since \( p \) and \( q \) do not end in zero, \( a \) and \( b \) must be at least 1.
4. We need to determine the last digit of \( p - q \). Since \( p = 5^a \) and \( q = 2^b \), we need to consider the last digits of these numbers:
   - The last digit of \( 5^a \) is always 5 for any positive integer \( a \).
   - The last digit of \( 2^b \) cycles through 2, 4, 8, and 6 as \( b \) increases.
5. We now consider the possible last digits of \( p - q \):
   - If \( q \equiv 2 \pmod{10} \), then \( p - q \equiv 5 - 2 \equiv 3 \pmod{10} \).
   - If \( q \equiv 4 \pmod{10} \), then \( p - q \equiv 5 - 4 \equiv 1 \pmod{10} \).
   - If \( q \equiv 8 \pmod{10} \), then \( p - q \equiv 5 - 8 \equiv -3 \equiv 7 \pmod{10} \).
   - If \( q \equiv 6 \pmod{10} \), then \( p - q \equiv 5 - 6 \equiv -1 \equiv 9 \pmod{10} \).
6. From the above calculations, the possible last digits of \( p - q \) are 1, 3, 7, and 9.
7. The only digit that does not appear in the list of possible last digits is 5.

The final answer is \(\boxed{5}\).