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) (Show Source): You can put this solution on YOUR website!
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) (Show Source): You can put this solution on YOUR website!
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
My calculator said it, I believe it, that settles it
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
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
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