The pth base ten repunit and the qth base ten repunit are respectively denoted by R(p) and R(q), where each of p and q exceeds 10.

Prove that R(p)*R(q) can never be equal to a palindrome.

Well, let's consider this as an exercise in long multiplication.

Consider if p & q were allowed to be less than 10. Let's say p = 3 and q = 4. Then R(p) * R(q)=

1 1 1

X 1 1 1 1

-----------

1 1 1

1 1 1

1 1 1

1 1 1

--------------

1 2 3 3 2 1 , which is palindromic, because no column exceeds 9 1's

If P and Q are > 10, then R(p)*R(q) must end in 987654321, but it cannot start with 123456789 because there are more than 9 1's in column 10 (from the left), so the addition has a carryover. and turns the starting 9 digits into 123456790.

Therefore, no palindrome.

Not clear why this is rated as difficulty 3. Am I missing something? I haven't actually carried out a multiplication where p and q > 10, but it seems obvious. In fact, I think p and q > 9 will suffice.