Answer:
The 3 digit numbers are;
171, 152, 133, 114, 285, 266, 247, 228, 209, 399
Explanation:
Let the 3 digit number be xyz
Since x is the Hundreds digit, y is the tens digit and z is the unit digit, we have;
100x, 10y and z
We are told the sum of this is equal to 19 times the sum of their digits.
Thus;
100x + 10y + z = 19(x + y + z)
100x + 10y + z = 19x + 19y + 19z
This gives;
81x - 9y - 18z = 0
Divide through by 9 to get;
9x - y - 2z = 0
x = (2z + y)/9
Now, we will need to select values of z and y between 0 to 9 in such a way that (2z + y) is divisible by 9 and in such a way that x > 0 but ≤ 10
The possible solutions are;
At z = 1, y = 7 and x = 1
At z = 2, y = 5 and x = 1
At z = 3, y = 3 and x = 1
At z = 4, y = 1 and x = 1
At z = 5, y = 8 and x = 2
At z = 6, y = 6 and x = 2
At z = 7, y = 4 and x = 2
At z = 8, y = 2 and x = 2
At z = 9, y = 0 and x = 2
At z = 9, y = 9 and x = 3
Compiling all these, the 3 digit numbers are;
171, 152, 133, 114, 285, 266, 247, 228, 209, 399