sum of the digits of a three-digit number is 20. If the hundreds and unit digits were interchanged, the resulting number is 297 less than the original. Find the number if the hundreds digit exceeds the tens digit by one.

let the unit digit be x

let the tens digit be y
let the hundred digit by (y+1)

y+1 + y + x = 20
2y+ x = 19
x = 19-2y, where both x and y must be whole numbers from 0 to 9.

the number is 100(y+1) + 10y + x
= x + 110y + 100

reverse the hundreds and unit digit:
100x + 10y + y+1
= 100x + 11y + 1

x + 110y + 100 - (100x + 11y + 1) = 297
110y+x+100-100x-11y-1 = 297
99y - 99x = 198
y - x = 2
y - 2(19-2y) = 2
3y=21
y=7

So our number is 875

check:
the digits of 875 add up to 20 ? YES
let's interchange the 8 and 5 ---> 578
what is the difference?
875-578 = 297 , all is good

Let's break down the problem step by step:

Step 1: Set up the equations for the given information.
Let's assume the three-digit number is represented as ABC, where A represents the hundreds digit, B represents the tens digit, and C represents the units digit.

The sum of the digits is given as A + B + C = 20 ------ Equation 1
The number obtained by interchanging the hundreds and units digits is 297 less than the original number, which can be represented as:
100C + 10B + A = 100A + 10B + C - 297 ----- Equation 2

Step 2: Simplify Equation 2.
Simplifying Equation 2, we get:
99C - 99A = 297
Dividing both sides by 99, we have:
C - A = 3 ------ Equation 3

Step 3: Utilize the information that the hundreds digit exceeds the tens digit by one.
Since A (the hundreds digit) exceeds B (the tens digit) by one, we have:
A - B = 1 ------ Equation 4

Step 4: Solve the system of equations.
Now we have two equations with two variables (Equations 3 and 4). Let's solve the system of equations.

Adding Equation 3 and Equation 4, we get:
(A - B) + (C - A) = 1 + 3
Simplifying, we have:
C - B = 4 ------ Equation 5

Using Equation 5, we can substitute C - B = 4 into Equation 1 to eliminate B:
A + B + C = 20
A + (C - 4) + C = 20
2A + 2C - 4 = 20
2A + 2C = 24
Dividing both sides by 2, we get:
A + C = 12

Now we have two equations: A + C = 12 ------ Equation 6 and C - A = 3 ------ Equation 3

Adding Equation 6 and Equation 3, we get:
(A + C) + (C - A) = 12 + 3
Simplifying, we have:
2C = 15
Dividing both sides by 2, we get:
C = 7

Substituting C = 7 into Equation 3, we have:
7 - A = 3
Adding A to both sides, we get:
7 = 3 + A
Simplifying, we have:
A = 4

Substituting A = 4 and C = 7 into Equation 6, we have:
4 + 7 = 12
Simplifying, we have:
11 = 12

This system of equations has no solution. Therefore, there is no three-digit number that satisfies all the given conditions.

To find the three-digit number, let's analyze the given information step by step.

Let's assume the three-digit number is represented as ABC, where A is the hundreds digit, B is the tens digit, and C is the units digit.

1. The sum of the digits of the three-digit number is 20. This gives us the equation: A + B + C = 20.

2. The hundreds and unit digits were interchanged, resulting in a number that is 297 less than the original. Swapping the digits gives us the number CBA. So, the equation becomes: 100C + 10B + A = 100A + 10B + C - 297.

3. The hundreds digit exceeds the tens digit by one. This gives us the equation: A = B + 1.

Now, we have a system of equations:
A + B + C = 20
100C + 10B + A = 100A + 10B + C - 297
A = B + 1

We can solve this system of equations to find the values of A, B, and C.

Using the third equation, we can rewrite the first equation as (B + 1) + B + C = 20, which simplifies to 2B + C = 19.

Now, substitute A = B + 1 in the second equation: 100C + 10B + (B + 1) = 100(B + 1) + 10B + C - 297.

Simplifying further, we get: 100C + 11B + 1 = 100B + 100 + 10B + C - 297.

Combining like terms: 100C + 11B + 1 = 110B - 197 + C.

Rearranging: 99C + 11B = 110B - 198.

Now, let's check the possible values for C from 0 to 9:

For C = 0, the equation becomes: 11B = 110B - 198.
Simplifying further, we get: 109B = 198. This equation has no integer solutions, so C cannot be 0.

For C = 1, the equation becomes: 99 + 11B = 110B - 198.
Simplifying further, we get: -11B + 110B = -297. By solving this, we find B = 9.

Finally, substitute the value of B into the first equation: A + 9 + 1 = 20.
Simplifying, we get A = 10 - 1, which means A = 9.

Therefore, the three-digit number can be represented as 9AB, which, in this case, is 919.