Let's define the three-digit number as:

$N = 100a + 10b + c$ where:

• $a$ is the hundreds digit,
• $b$ is the tens digit,
• $c$ is the ones digit.

From the problem, we know:

1. The sum of the digits is 17: a + b + c = 17 \tag{1}

2. When the digits are reversed, the new number becomes $100c + 10b + a$, and when this reversed number is added to the original number, the result is 1,474: (100a + 10b + c) + (100c + 10b + a) = 1474 \tag{2} Simplifying the equation: 101a + 20b + 101c = 1474 \tag{3}

3. When the reversed number is subtracted from the original number, the result is 396: (100a + 10b + c) - (100c + 10b + a) = 396 \tag{4} Simplifying the equation: 99a - 99c = 396 \tag{5} Dividing by 99: a - c = 4 \tag{6}

Now, we have the following system of equations:

1. $a + b + c = 17$
2. $101a + 20b + 101c = 1474$
3. $a - c = 4$

Step 1: Solve for $a$ and $c$

From equation (6), we know: $a = c + 4$

Step 2: Substitute $a = c + 4$ into equation (1)

Substitute $a = c + 4$ into the sum of digits equation: $(c + 4) + b + c = 17$ Simplify: $2c + b + 4 = 17$ 2c + b = 13 \tag{7}

Step 3: Substitute $a = c + 4$ into equation (3)

Substitute $a = c + 4$ into the equation for the sum of the original and reversed numbers: $101(c + 4) + 20b + 101c = 1474$ Simplify: $101c + 404 + 20b + 101c = 1474$ $202c + 20b + 404 = 1474$ Subtract 404 from both sides: 202c + 20b = 1070 \tag{8}

Step 4: Solve the system of equations

Now we have the system of two equations:

1. $2c + b = 13$
2. $202c + 20b = 1070$

We can solve this system step-by-step:

• From equation (7), solve for $b$: $b = 13 - 2c$

• Substitute this expression for $b$ into equation (8): $202c + 20(13 - 2c) = 1070$ Simplify: $202c + 260 - 40c = 1070$ Combine like terms: $162c + 260 = 1070$ Subtract 260 from both sides: $162c = 800$ Divide by 162: $c = \frac{800}{162} = 4.938$ Since $c$ must be an integer, the result doesn't match a valid digit.