<PRE><B>Question 1187151</B>
&lt;br&gt;
Given the two 3-digit numbers ABC and CBA, and assuming A is greater than C, then the difference ABC-CBA is equal to 99 times (A-C).&lt;br&gt;
Since that difference is 396, we know A-C is 396/99 = 4.&lt;br&gt;
So the hundreds digit is 4 more than the units digit; and the problem says the tens digit is 0 and the sum of the other two digits is 6.  Then we have&lt;br&gt;
A-C = 4
A+C = 6&lt;br&gt;
which gives us A=5 and C=1.&lt;br&gt;
ANSWER: The original number is 501&lt;br&gt;

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