Question 23556
Let the hundreds, tens, and units digits be H, T, and U, respectively
Then H + T + U = 19 -------- eq (i)


Now, since the hundreds digit is 3 times the tens digit, then H = 3T
Substituting 3T for H in eq (i), we can say that 3T + T + U = 19, or 4T + U = 19 ------ eq (ii) 


Original number’s value = 100(3T) + (10T) + U ----> 300T + 10T + U ----> 310T + U

Reversed number’s value = 100(U) + 10(T) + 3T ----> 100U + 10T + 3T ----> 100U + 13T


Since the original number is 198 more than when it’s reversed, then we’ll have: 

Original number = Reversed number + 198, or


310T + U = 100U + 13T + 198 (notice that the middle digit remains the same although the 3-digit number has been reversed)


310T – 13T – 99U = 198 
297T – 99U = 198 ------ eq (iii)
396T + 99U = 1881 ------ eq (iv)------ Multiply eq (ii) by 99
693T  =  2,079 ---- Adding eqs (iii) and (iv)
T = 3


Substitute 3 for T in eq (ii) to get:  4(3) + U = 19………… U = 7

Substitute 7 for U in eq (i) to get: H + 3 + 7 = 19 ……….. H = 9


H = 9, T = 3, and U = 7. Therefore, the original number is {{{highlight_green(937)}}}. When reversed, it becomes 739, which is 198 less than the original number