Consider a three digit number ABC, where the letters represent digits and not factors. If C&gt;A, we can simply break this  number into ABA + [C-A]. If C=A, it is already a palindrome. If C&lt;A, we can still split it apart as before as long as we can borrow 1 from B. So B&gt;0. For example, 381 becomes 373 + 8.<br>
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If B=0 and C&gt;A, ABC (or A0C) becomes A0A + C. If C&lt;A, then A0C becomes [A-1]9[A-1] + (10 + C - (A-1)). As long as 10+C-A+1  or 11+C-A is less than 10, or<br>
11+C-A&lt;10<br>
C-A&lt;-1<br>
C&lt;A-1,<br>
then sums of the form ABA + C can be obtained. Numbers that don't satisfy this are of the form A0[A-1]. They include<br>
908, 807, 706, 605, 504, 403, 302, 201, 100.<br>
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100, we observe, is 99 + 1, so this can be eliminated.<br>
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To try and sum these, we try 3-digit plus 2-digit palindromes, or A0[A-1] = [A-1][10-N][A-1-N] + NN, where N is a digit. 10-N will always be a digit, but we need A-1 = A-1-N. This is impossible.<br>
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Our only option is 3-digit plus 3-digit. So A0[A-1] = [A-P-1][10-Q][A-1-P] + PQP, which works if A-1-P is a digit. P is at least 1, and ideally, exactly 1. For instance, 908 = 7[10-Q]7 + 1Q1. Any digit for Q works. 757 + 151, 767 + 141, 777 +  131, etc. All sum to 908.<br>
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Down to 302, A-1-P is a digit. But in the case of 201, A-1-P is 0 if P is chosen to be 1. Therefore <strong>201</strong> cannot be expressed as the sum of two palindromes.