Question 887638
a 3-digit number is 99 more than itself reversed.
<pre>
      100h+10t+u = 100u+10t+h+99

Simplify

        99h-99u = 99

Divide through by 99

            h-u = 1
</pre>
the units digit exceeds the tens digit by 2
<pre>
u=t+2
</pre>
the hundreds digit exceeds the unit digit by 1
<pre>
h=u+1

So we have to solve this system:

{{{system(h-u=1,u=t+2,h=u+1)}}}

h  -u=1
 -t+u=2  
h  -u=1

The augmented matrix is 

{{{(matrix(3,5,
1,0,-1,"|",1,
0,-1,1,"|",2,
1,0,-1,"|",1))}}}

Its row reduced echelon form is 

{{{(matrix(3,5,
1,0,-1,"|",1,
0,1,-1,"|",-2,
0,0,0,"|",0))}}}

Which becomes the system

h  -u= 1 
  t-u=-2

h = u+1
t = u-2

So we can choose u as 2 through 8,
Seven solutions:

u=_, h=u+1=_, t=u-2=_, number=htu                              htu
----------------------------------
u=2, h=2+1=3, t=2-2=0, number=302
u=3, h=3+1=4, t=3-2=1, number=413
u=4, h=4+1=5, t=4-2=2, number=524
u=5, h=5+1=6, t=5-2=3, number=635
u=6, h=6+1=7, t=6-2=4, number=746
u=7, h=7+1=8, t=7-2=5, number=857
u=8, h=8+1=9, t=8-2=6, number=968

Edwin</pre>