All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Repunit Product, Palindrome Not (Posted on 2009-09-30)
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.

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Not very formal, but ... (spoiler) | Comment 1 of 2
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.

 Posted by Steve Herman on 2009-10-01 01:20:32

 Search: Search body:
Forums (0)