For any three digit number.

Let

x = the hundreds digit

y = the tens digit and

z = the units digit

The number = 100x + 10y + z

The number with digits reversed = 100z + 10y + x

The sum of digits = x + y + z

The product of digits = xyz

Example

In a three digit number, the hundreds digit is twice the units digit. If 396 be subtracted from the number, the order of the digits will be reversed. Find the number if the sum of the digits is 17.

Solution

$x$ = hundreds digit

$y$ = tens digit

$z$ = units digit

$100x + 10y + z$ = the number

The hundreds digit is twice the units digit

$x = 2z$ → equation (1)

The sum of the digits is 17

$x + y + z = 17$ → equation (2)

396 be subtracted from the number

$(100x + 10y + z) - 396 = 100z + 10y + x$

$99x - 99z = 396$

$x - z = 4$ → equation (3)

Substitute x = 2z to equation (3)

$2z - z = 4$

$z = 4$

From equation (1)

$x = 2(4)$

$x = 8$

From equation (2)

$8 + y + 4 = 17$

$y = 5$

The number is

$100x + 10y + z = 854$ *answer*