SOLUTION: Find the largest 5-digit palindrome that is divisoble by 101.

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Question 620548: Find the largest 5-digit palindrome that is divisoble by 101.
Found 3 solutions by Alan3354, solver91311, richard1234:
Answer by Alan3354(69443) About Me  (Show Source):
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Find the largest 5-digit palindrome that is divisoble by 101.
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40804 is the largest I know.

Answer by solver91311(24713) About Me  (Show Source):
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We need to find a 3 digit number ABC such that ABC times 101 is a five-digit palindrome.

In order for the product to be a palindrome given that one of the factors (101) is a palindrome, the other factor must be a palindrome also, hence we are actually looking for a three digit factor ABA.






And the product will be a palindrome if and only if there is no carry when you add A + A. Hence, the largest A can be is 4. Find the largest 3 digit palindrome where the first and last digit is 4, then multiply that times 101 to find your largest 5 digit palindrome.

John

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Answer by richard1234(7193) About Me  (Show Source):
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Suppose our palindrome is abcba, where a,b,c are digits, and





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



Modulo 101, and . Therefore,

. 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

49894