a 3-digit number is 99 more than itself reversed.
100h+10t+u = 100u+10t+h+99
Simplify
99h-99u = 99
Divide through by 99
h-u = 1
the units digit exceeds the tens digit by 2
u=t+2
the hundreds digit exceeds the unit digit by 1
h=u+1
So we have to solve this system:
h -u=1
-t+u=2
h -u=1
The augmented matrix is
Its row reduced echelon form is
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
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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
