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Math Help - Palindromes

  1. #1
    Member rtblue's Avatar
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    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:

    [100a+10b+a][100c+10d+c]=436995

    [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.
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  2. #2
    MHF Contributor harish21's Avatar
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    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 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!
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  3. #3
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    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: . 3^4\cdot5\cdot13\cdot83

    . . and found two palindromic factors: . 585 and 747.


    Their sum is 1332.

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