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?

2 Answers
Aug 6, 2016

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

Aug 6, 2016

=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