Example 01 | Digit-related problem

Let
$x$ = the hundreds digit
$y$ = the tens digit
$z$ = the units digit
 

$100x + 10y + z$       → the original number

$100x + 10z + y$       → the tens and units digits are interchanged

$100y + 10x + z$       → the hundreds and tens digits are interchanged
 

The sum of the digits of a three-place number is 19
$x + y + z = 19$       → Equation (1)
 

If the tens and units digits are interchanged the number is diminished by 27
$(100x + 10z + y) = (100x + 10y + z) - 27$

$-9y + 9z = -27$

$-y + z = -3$

$z = y - 3$       → Equation (2)
 

If the hundreds and tens digits are interchanged the number is increased by 180
$(100y + 10x + z) = (100x + 10y + z) + 180$

$90y - 90x = 180$

$y - x = 2$

$y = x + 2$       → Equation (3)
 

Substitute y = x + 2 to Equation (2)
$z = (x + 2) - 3$

$z = x - 1$
 

Substitute y = x + 2 and z = x - 1 to Equation (1)
$x + (x + 2) + (x - 1) = 19$

$3x = 18$

$x = 6$

From Equation (3)
$y = 6 + 2$

$y = 8$

From Equation (2)
$z = 8 - 3$

$z = 5$
 

The number is
$100x + 10y + z = 685$           answer