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| Math Central | Quandaries & Queries |
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Question from pradip, a student: I have discovered that for any n-digit number, when it's terminal digits are reversed,e.g.,1729 & 9721 & the resulting combinations are subtracted, the resulting solution is always a multiple of 9 & when this multiple of 9 is divided by 9, we will always get a palindrome, Example- 1234 & 4231, 4231-1234=2997/9=333., please I am not sure about the above fact, would please help me in confirming my discovery. |
Pradip,
It looks to me as if you are taking four digits a, b, c and d, constructing two numbers abcd and dbca, and then subtracting them. If this is correct then the numbers you have are
a × 103 + b × 102 + c × 10 + d and d × 103 + b × 102 + c × 10 + a
Subtracting the second from the first gives ( I am assuming a is larger than d so that the subtraction yields a positive number.)
(a × 103 + b × 102 + c × 10 + d) - (d × 103 + b × 102 + c × 10 + a)
= (a - d) × 103 + (d - a)
= (a - d) × (103 - 1)
= (a - d) × (1000 - 1)
= (a - d) × 999
= 9 (a - d) × 111
Division by 9 yields
(a - d) × 111
This is the digit a - d, repeated three times.
A similar result will occur is you start with any positive integer with an even number of digits. I don't know how you are treating a positive integer with an odd number of digits.
Harley
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