Question

A two digit number is seven times the sum of it's digits. The number formed by reversing the digits is 6 more than half of the original number. Find the difference of the digits of the given number .

Answer 1

a = first number

b = second number

Your number = 10 a + b

A two digit number is seven times the sum of it's digits meaans:

10 a + b = 7 ( a + b )

The number formed by reversing the digits is 10 b + a

The number formed by reversing the digits is 6 more than half of the original number means:

10 b + a = ( 10 a + b ) / 2 + 6

Now you must solve system of two equations:

10 a + b = 7 ( a + b )

10 b + a = ( 10 a + b ) / 2 + 6
____________________________

First equation:

10 a + b = 7 a + 7 b

Subtract 7 a to both sides

3 a + b = 7 b

Subtract b to both sides

3 a = 6 b

Divide both sides by 3

a = 2 b

Replace a with 2 b in equation:

10 b + a = ( 10 a + b ) / 2 + 6

10 b + 2 b = ( 10 • 2 b + b ) / 2 + 6

12 b = ( 20 b + b ) / 2 + 6

12 b = 21 b / 2 + 6

Multiply both sides by 2

24 b = 21 b + 12

Subtract 21 b to both sides

3 b = 12

Divide both sides by 3

b = 4

a = 2 b

a = 2 • 4

a = 8

a - b = 8 - 4 = 4

Check result:

A two digit number is seven times the sum of it's digits.

Your number = 10 a + b = 10 • 8 + 4 = 84

Sum of digits = 8 + 4 = 12

84 = 7 •12

The number formed by reversing the digits is 6 more than half of the original number.

The number formed by reversing the digits is 10 b + a = 10 • 4 + 8 = 48

Half of the original number =

84 / 2 = 42

48 = 42 + 6

Answer 2

a = first digit

b = second digit