Question 79332
I'll use a for 100's digit,
b for the 10's disit
c for the ones digit
{{{a + b + c = 9}}}
and the problem also states that
{{{a + b = c - 1}}}
Substitute {{{a + b}}} in the 2nd for {{{a + b}}} in the 1st
{{{c - 1 + c = 9}}}
{{{2c = 10}}}
{{{c = 5}}}
And the problem states that
{{{(100a + 10b + c) / 9 = 3c}}}
{{{100a + 10b + c = 27c}}}
{{{100a + 10b = 26c}}}
{{{50a + 5b = 13c}}}
Note that if  {{{c = 5}}} and {{{a + b = c - 1}}}, then
{{{a + b = 4}}}
{{{b = 4 - a}}}
Substitute for b
{{{50a + 5(4 - a) = 13*5}}}
{{{50a + 20 - 5a = 65}}}
{{{45a = 45}}}
{{{a = 1}}}
{{{b = 4 - a}}}
{{{b = 3}}}
Now I have a,b and c.
The number is {{{135}}}
check
Does {{{a + b + c = 9}}}? yes
Does {{{(100a + 10b + c) / 9 = 3c}}} yes