1. ## word problem

"The sum of the first two digits of a three-digit number exceeds the third digit by 8. If the first two digits are interchanged the new number exceeds the original number by 270. If the number is divided by the sum of the digits the quotient is 31 and the remainder 15. Find the original number."

2. Originally Posted by biva
"The sum of the first two digits of a three-digit number exceeds the third digit by 8. If the first two digits are interchanged the new number exceeds the original number by 270. If the number is divided by the sum of the digits the quotient is 31 and the remainder 15. Find the original number."
Let the number be "xyz." That is to say the number is
$\displaystyle 100x + 10y + z$

We know that
$\displaystyle x + y = z + 8$

$\displaystyle 100y + 10x + z = (100x + 10y + z) + 270$

$\displaystyle \frac{100x + 10y + z}{x + y + z} = 31 + \frac{15}{x + y + z}$

$\displaystyle z = x + y - 8$

Inserting this into the other two expressions gives:
$\displaystyle 100y + 10x + (x + y - 8) = (100x + 10y + (x + y - 8)) + 270$ ==> $\displaystyle 101y + 11x - 8 = 101x + 11y + 262$

and
$\displaystyle \frac{100x + 10y + (x + y - 8)}{x + y + (x + y - 8)} = 31 + \frac{15}{x + y + (x + y - 8)}$ ==> $\displaystyle \frac{101x + 11y - 8}{2x + 2y - 8} = 31 + \frac{15}{2x + 2y - 8}$

Let's look at the new top equation:
$\displaystyle 101y + 11x - 8 = 101x + 11y + 262$

$\displaystyle -90x + 90y = 270$

$\displaystyle 90y = 90x + 270$

$\displaystyle y = x + 3$

Inserting this into the new bottom equation:
$\displaystyle \frac{101x + 11(x + 3) - 8}{2x + 2(x + 3) - 8} = 31 + \frac{15}{2x + 2(x + 3) - 8}$

$\displaystyle \frac{112x + 25}{4x - 2} = 31 + \frac{15}{4x - 2}$<-- Multiply both sides by 4x - 2

$\displaystyle 112x + 25 = 31(4x - 2) + 15$

$\displaystyle 112x + 25 = 124x - 62 + 15$

$\displaystyle -12x + 25 = - 47$

$\displaystyle -12x = -72$

$\displaystyle x = 6$

So
$\displaystyle y = x + 3$ ==> $\displaystyle y = 9$

and

$\displaystyle z = x + y - 8$ ==> $\displaystyle z = 7$

-Dan

3. Hello, biva!

Another approach . . . with fewer fractions.

The sum of the first two digits of a three-digit number exceeds the third digit by 8.
If the first two digits are interchanged, the new number exceeds the original number by 270.
If the number is divided by the sum of the digits, the quotient is 31 and the remainder 15.
Find the original number.
Let the digits be $\displaystyle x,\,y,\,z$.
The three-digit number is: $\displaystyle 100x + 10y + z$

The first sentence says: .$\displaystyle x + y \:=\:z + 8\quad\Rightarrow\quad x + y - z \:=\:8$ .[1]

The second says: .$\displaystyle 100y + 10x + z \:=\:100x+ 10y + z + 270\quad\Rightarrow\quad x - y \:=\:-3$ .[2]

The third says: .$\displaystyle \frac{100x + 10y + z}{x + y + z} \:=\:31 + \frac{15}{x + y + z}\quad\Rightarrow\quad23x - 7y - 10z \:=\:5$ .[3]

We have a system of equations.

From [2], we have: .$\displaystyle y \:=\:x + 3$ .[4]

Substitute into [1]: .$\displaystyle x + (x + 3) - z \:=\:8\quad\Rightarrow\quad 2x - z\:=\:5$ .[5]

Substitute into [3]: .$\displaystyle 23x - 7(x+3) - 10z \:=\:5\quad\Rightarrow\quad 8x - 5z \:=\:13$ .[6]

Multiply [5] by -5: .$\displaystyle \text{-}10x + 5x \:=\:\text{-}25$
. . . . . . .Add [6]: . . $\displaystyle 8x - 5x \:=\;13$

. . and we get: .$\displaystyle -2x \:=\:-12\quad\Rightarrow\quad\boxed{x \,=\,6}$

Substitute into [4]: .$\displaystyle y \:=\:6 + 3\quad\Rightarrow\quad\boxed{y \,=\,9}$

Substitute into [5]: .$\displaystyle 2(6) - z \:=\:5\quad\Rightarrow\quad\boxed{z \,=\,7}$

Therefore, the original number is
697
.