Palindrome Me!
and are four-digit positive integers. To maximize the difference between the two positive integers, what is the smallest possible value of ?
Note: and cannot be 0.
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1 solution
Value of ABCD = 1000A + 100B + 10C + D
Value of DCBA = 1000D + 100C + 10B + A
Subtracting leaves 999A + 90B - 90C - 999D
Rearranging and simplifying gives 999( A - D) + 90( B - C)
To maximize the value A - D ( neither can be 0 ) they must be 9 and 1
To maximize the value B - C they must be 9 and 0
So 9901 - 1099 = 8802
Each of the other values subtracted from their palindrome gives 7992 or 8712
Therefore smallest of the two numbers is 1099 which is not on the list.