Question 1033057
<font face="Times New Roman" size="+2">


You only get to ask one question per post, and since your first question is nonsense, I'll do the second one.  If you are having trouble with English, get someone to help you.


Let *[tex \Large x] represent the 100s digit of the 3-digit number, *[tex \Large y] represent the 10s digit, and *[tex \Large z] represent the 1s digit.  Then *[tex \Large 100x\ +\ 10y\ +\ z] is a representation of the original number.  The number obtained when the digits are reversed is then *[tex \Large 100z\ +\ 10y\ +\ x].  And their difference is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 100x\ +\ 10y\ +\ z\ -\ (100z\ +\ 10y\ +\ x)\ =\ 99x\ -\ 99z]


Which is evenly divisible by 99, hence the remainder is zero.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

</font>