1. ## Palindromes

The product of two positive three-digit palindromes is 436995. What is their sum?

By the way, a palindrome is number that is read the same backwards and forwards: example: 1331, 7557, 131, 767, 454, etc..

This is what I have tried:

$\displaystyle [100a+10b+a][100c+10d+c]=436995$

$\displaystyle [101a+10b][101c+10d]=436995$

I realize that either a or c must be 5, for the product to end in a 5. Also I see that one or both of these numbers must be a factor of 9 and 3, but i'm not sure what to make of these observations.

I'm not exactly sure where to go from here. Help is appreciated.

2. Here's a simple method:

since the product of the two numbers is 436,995 (which ends with 5), one of the two numbers should end with 5.And since it is a palindrome, it should also begin with 5. So, this number is is $\displaystyle 5a5$ ;where a can be any number from 0 to 9.

Now you can divide the number 436,995 by 5a5 till the result get a palindrome..

436,995/505= 865.33

436,995/515=848.53

436,995/525=832.37

436,995/535=816.81

436,995/545=801.82

436,995/555=787.37

436,995/565=773.44

436,995/575=759.99

436,995/585=747 ...There you go!

3. Hello, rtblue!

The product of two positive three-digit palindromes is 436995.
What is their sum?

I found no satisfactory algebraic approach.

So I factored 436,995: .$\displaystyle 3^4\cdot5\cdot13\cdot83$

. . and found two palindromic factors: .$\displaystyle 585$ and $\displaystyle 747.$

Their sum is $\displaystyle 1332.$