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Consecutive Near Palindromes (Posted on 2008-02-13) Difficulty: 3 of 5
Find a set of four consecutive six digit numbers which satisfies the following:
  • The last four digits of the first (smallest) number form a palindrome.
  • The last five digits of the second number form a palindrome.
  • The second through fifth digits of the third number form a palindrome.
  • The fourth (largest) number is a palindrome.

See The Solution Submitted by Brian Smith    
Rating: 3.3333 (3 votes)

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Solution solution (spoiler) | Comment 1 of 8

The fourth (largest) number is of the form

ABCCBA

To get the third number you subtract 1. In doing so, you can't do a "borrow" as that would upset the BCCB palindrome that is said to exist in that third number.


Working back another, the second number is supposed to be palindromic in the last 5 digits, so by this time, there has been a "borrow" and the B has changed to a C, and the A is now a B:

ABCCCB

so the A's must be 1's and the B's must be 9's, and the C must be one less than the B, and therefore are 8's.

The largest number is therefore

198891

and the earlier numbers follow:

198890
198889
198888


  Posted by Charlie on 2008-02-13 11:31:25
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