# Example 01 | Digit-related problem

Problem

The sum of the digits of a three-place number is 19. If the tens and units digits are interchanged the number is diminished by 27, and if the hundreds and tens digits are interchanged the number is increased by 180. What is the number?

Solution

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$x$ = 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*

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