Question

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?

Answer 1

53

Explanation:

Number with two digits can be expressed as :

`10n_(2) + n_(1)` for `n_1, n_2 in ZZ`

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 `10n_(2) + n_(1)` while the unit digit will be `n_1`.

`10n_(2) + n_(1) = 17n_1 + 2`

`therefore 10n_2 - 16n_1 = 2`

Substituting:

`10(8-n_1) - 16n_1 = 2`

`80 - 26n_1 = 2`

`26n_1 = 78 implies n_1 = 3`

`n_2 = 8 - n_1 = 8 - 3 = 5`

`therefore` number is `53`

Answer 2

`=53`

Explanation:

Let the unit digit be `y` and ten-digit be `x`
So the number is `10x+y`
So we get
`x+y=8` and
`10x+y=17y+2`
or
`10x+y-17y=2`
or
`10x-16y=2`
Dividing both the sides by 2 we get
`5x-8y=1` From the equation `x+y=8` we get 8x+8y=64
Adding up we get
`5x-8y+8x+8y=64+1`
or
`5xcancel(-8y)+8xcancel(+8y)=65`
or
`13x=65`
or
`x=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 `10x+y=10(5)+3=53`