# The sum of the digits of a two number is 8. The number exceeds 17 times the unit’s digit by 2. How do you find the number?

Aug 6, 2016

53

#### Explanation:

Number with two digits can be expressed as :

$10 {n}_{2} + {n}_{1}$ for ${n}_{1} , {n}_{2} \in \mathbb{Z}$

We know that the sum of the two digits is 8 so:

${n}_{1} + {n}_{2} = 8 \implies {n}_{2} = 8 - {n}_{1}$

The number is 2 more than 17 times the unit digit. We know that the number is expressed as $10 {n}_{2} + {n}_{1}$ while the unit digit will be ${n}_{1}$.

$10 {n}_{2} + {n}_{1} = 17 {n}_{1} + 2$

$\therefore 10 {n}_{2} - 16 {n}_{1} = 2$

Substituting:

$10 \left(8 - {n}_{1}\right) - 16 {n}_{1} = 2$

$80 - 26 {n}_{1} = 2$

$26 {n}_{1} = 78 \implies {n}_{1} = 3$

${n}_{2} = 8 - {n}_{1} = 8 - 3 = 5$

$\therefore$ number is $53$

Aug 6, 2016

$= 53$

#### Explanation:

Let the unit digit be $y$ and ten-digit be $x$
So the number is $10 x + y$
So we get
$x + y = 8$ and
$10 x + y = 17 y + 2$
or
$10 x + y - 17 y = 2$
or
$10 x - 16 y = 2$
Dividing both the sides by 2 we get
$5 x - 8 y = 1$ From the equation $x + y = 8$ we get 8x+8y=64
$5 x - 8 y + 8 x + 8 y = 64 + 1$
or
$5 x \cancel{- 8 y} + 8 x \cancel{+ 8 y} = 65$
or
$13 x = 65$
or
$x = \frac{65}{13}$
or
$x = 5$

By putting the value $x = 5$ in $x + y = 8$
we get
$5 + y = 8$
or
$y = 8 - 5$
or
$y = 3$
Hence the number is $10 x + y = 10 \left(5\right) + 3 = 53$