Question 561896
If a,b,c denote the digits, then


*[tex \LARGE 100a + 10b + c = 17a + 34b + 51c] (right away we know the number is a multiple of 17)


*[tex \LARGE 83a = 24b + 50c]


Here, we conclude that a is even (since RHS is even). Now consider the equation mod 10 (if you don't know modular arithmetic, think of it as "matching the units digits on LHS and RHS").


If a = 2, then b = 4 or 9, but neither allows for an integer value of c. If a = 4, then b = 3 or 8, but likewise, there are no integer solutions for c. If a = 6, then b = 2 or 7, and it can be checked that b = 2 allows for the number 629. If a = 8, then b = 1 or 6, but neither allows for integer solutions.


Hence the only number is 629.