Question

The sum of the digits of a three-place number is 19. If the tens and units digits are interchange the number is diminished by 27, and if the hundreds and tens digits are interchange the number is increased by 180. What is the number? Set up sum of three didgits with only 2 variables.

Answer 1

To solve this problem, we can use variables to represent each digit of the three-place number and set up a system of equations. By solving the system of equations, we can find the values of the variables and determine the number. The number is 109.

Step-by-step explanation:

To solve this problem, let's use a variable to represent each digit of the three-place number. Let the hundreds digit be represented by 'x,', the tens digit by 'y,' and the ones digit by 'z.' The given conditions can be expressed as the following equations:

1. x + y + z = 19
2. 100z + 10y + x - 100x - 10y - z = -27
3. 100y + 10z + x - 100x - 10z - y = 180

From these equations, we can simplify by combining like terms:

1. 2x + 2y + 2z = 19
2. -99x + 9y + 99z = -27
3. -99x - 9y + 99z = 180

Now, we can eliminate the variable 'x' by adding equations 2 and 3:

1. -99x + 9y + 99z = -27
2. -99x - 9y + 99z = 180
3. -----------------------------
4. 0x + 0y + 198z = 153

Dividing both sides by 198, we find that 'z' = 153/198. Since we are dealing with a three-place number, 'z' can only be an integer value. The nearest integer to 153/198 is 1. Substituting 'z' = 1 back into equation 1, we can solve for 'y' and find 'y' = 8. Finally, we substitute the known values of 'y' and 'z' into equation 1 to find 'x' = 10 - 'y' - 'z' = 10 - 8 - 1 = 1.

Therefore, the number is 100x + 10y + z = 100(1) + 10(8) + 1 = 108 + 1 = 109.

Answer 2

I cant really help you out with the last part since setting up the sum of three digits with two variable complicates things but here is the calculation and answer as follows:

let the number be a be represented as xyz (where x is the hundredth digit, y is the tenth digit and z is the one digit)

which means a = 100x+10y+z (equation 1)
if you add the three digits x+y+z = 19 (equation 2)

if you switch positions of the tenth digit and one digit then
a-27 = 100x + 10z +y (equation 3) (number diminish by 27 hence why a -27)

if you switch the position of the hundredth and tenth digit then
a+180 = 100y + 10x + z (equation 4) (number increases by 180)

if you use equation 1, 2, 3, and 4 and solve for x, y and z which gives you a then the number is 685.

Sorry for the long explanation but this was the first approach that came to mind might be more simpler ones and also one that helps you set up the sum of three digits with only 2 variables.