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Count the palindromes (Posted on 2015-11-11) Difficulty: 2 of 5
N is both a a palindrome and a sum of eleven consecutive positive integers.

How many possible values for N exist below 80,000?
Please provide the lowest and the highest samples.

No Solution Yet Submitted by Ady TZIDON    
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Solution re(4): Simplifying the question Comment 7 of 7 |
(In reply to re(3): Simplifying the question by Ady TZIDON)

ANALYTICAL SOLUTION:


Any multiple of 11 starting with 66 works, if it is a palindrome

Two digits:  66, 77, 88, 99 -- total of 4

Three digits: 
  mod 11, 0 = aba = 101*a + b*10 = 2a - b
          b = 2a mod 11 has single digit solutions for all a except a = 5
          Total solutions = 9 - 1 = 8

Four digits:
  mod 11, 0 = abba = 1001*a + b*110 = 0
            all a and b work
          Total solutions = 90 (ie, 10 to 99)

Five digits:
  mod 11, 0 = abcba = 10001*a + b*1010 + 100c = 2a - 2b + c 
          c = 2(b-a) mod 11 has single digit solutions as long as (b-a) <> 5 mod 11
          ab under 80 that do not work are 16, 27, 38, 49, 60, 71
          Total solutions under 8 = 70 - 6 = 64
          
Grand total = 4 + 8 + 90 + 64 = 166, which has the advantage of agreeing with Charlie's computer  

  Posted by Steve Herman on 2015-11-12 12:33:21
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