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Question
The sum of a two-digit number and the number formed by reversing the order of digit is 66. If the two digits differ by 2, find the number. How many such numbers are there?
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Solution
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is ` 10 y + x`
The two digits of the number are differing by 2. Thus, we have ` x - y =+- 2`
After interchanging the digits, the number becomes `10 x + y.`
The sum of the numbers obtained by interchanging the digits and the original number is 66. Thus, we have
`(10 x + y ) + ( 10 y + x )= 66`
` ⇒ 10 x + y + 10 y + x = 66`
`⇒ 11 x + 11y = 66 `
` ⇒ 11 ( x + y) = 66/11`
` ⇒ x + y = 66/11`
` ⇒ x + y =6`
So, we have two systems of simultaneous equations
`x - y = 2`
` x + y = 6`
` x - y = -2`
` x + y = 6`
Here x and y are unknowns. We have to solve the above systems of equations for x and y.
(i) First, we solve the system
` x - y = 2`
` x + y = 6`
Adding the two equations, we have
`( x - y ) + ( x + y )= 2 + 6`
`⇒ x - y + x + y = 8`
` ⇒ 2x = 8`
` ⇒ x = 8/2`
` ⇒ x = 4`
Substituting the value of x in the first equation, we have
` 4 - y = 2`
`⇒ y = 4 -2`
`⇒ y = 2`
Hence, the number is ` 10 xx 2 + 4 = 24`.
(ii) Now, we solve the system
` x - y= -2`
` x + y = 6`
Adding the two equations, we have
` ( x - y) + ( x + y) = -2 + 6`
`⇒ x - y + x + y = 4 `
` ⇒ 2x = 4`
` ⇒ x = 4/2`
` ⇒ x = 2`
Substituting the value of x in the first equation, we have
` 2 - y = -2`
` ⇒ y = 2 + 2`
` ⇒ y = 4`
Hence, the number is ` 10 xx 4 + 2 = 42`
There are two such numbers.
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