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Question
A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.
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 number is 4 times the sum of the two digits. Thus, we have
` 10 y + x = 4 ( x + y)`
` ⇒ 10 y + x = 4x + 4 y `
`⇒ 4 x + 4 y - 10 y - x = 0 `
`⇒ 3 x - 6 y = 0 `
`⇒ 3( x - 2y)= 0`
` ⇒ x - 2y =0`
After interchanging the digits, the number becomes `10x + y .`.
If 18 is added to the number, the digits are reversed. Thus, we have
`(10 y + x )+ 18 =10x + y`
`⇒ 10 x + y -10y -x =18`
` ⇒ 9x -9y =18`
` ⇒ 9(x -y) = 18`
` ⇒ x - y= 18/9`
`⇒ x - y = 2`
So, we have the systems of equations
`x - 2y = 0`
`x - y = 2 `
Here x and y are unknowns. We have to solve the above systems of equations for xand y.
Subtracting the first equation from the second, we have
`( x - y)-(x - 2y)=2-0`
`⇒ x - y-x+2y=2 `
` ⇒ y = 2`
Substituting the value of y in the first equation, we have
` x - 2 xx2=0`
` ⇒ x - 4 =0`
` ⇒ x = 4`
Hence, the number is ` 10 xx2 + 4 = 24`.
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