Question 1132364
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<U>ANSWER</U>.  The maximal such a number is 399.



<U>Solution</U>


<pre>
Let "a" be the "hundredth" digit; "b" be the "tens" digit and "c" be the "ones" digit.


Then the number is  100a + 10b + c,  and the condition says


    100a + 10b + c = 19*(a+b+c),     or


    81a  - 9b - 18c = 0,


     9a  - b - 2c = 0.


Then  a= 3, b= 9  and c= 9   is 

    a) the solution to the last equation,    and 

    b) represents/provides the maximal such a number.


It is obvious, taking into account that  0 <= a <= 9,  0 <= b<= 9,  0 <= c <= 9.
</pre>

Solved.