Question 620548
Suppose our palindrome is abcba, where a,b,c are digits, and


*[tex \LARGE 101 | 10000a + 1000b + 100c + 10b + a]


*[tex \LARGE 101 | 10001a + 1010b + 100c]


1010b is already divisible by 101, so we can say that the above expression is equivalent to


*[tex \LARGE 101 | 10001a + 100c]


Modulo 101, *[tex \LARGE 10001 \equiv 2] and *[tex \LARGE 100 \equiv -1]. Therefore,


*[tex \LARGE 2a - c \equiv 0 (mod 101)]. Optimize and let a = 4, c = 8. Our choice of b doesn't matter, because 1010 is already 0 mod 101. Therefore, let b = 9. The largest palindrome multiple of 101 is


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