<PRE><B>Question 724253</B>
 let the digits of a three digit number be x,y and z. Then the value of the number is 100x+ 10y + z. Show that the difference between the number and reversed number is 99
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reversed number: 100z + 10y + x
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Equation:

(100x+ 10y + z) - (100z + 10y + x) = 99x - 99z = (99)(x-z)
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Example:
586 - 685 = -101
200-002 = 198
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Note: The claim made is not true as stated.
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Cheers,
Stan H.</PRE>