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
