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Question
A two-digit number is 4 more than 6 times the sum of its digits. If 18 is subtracted from 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 `10y+x.`.
The number is 4 more than 6 times the sum of the two digits. Thus, we have
` 10 y + x = 6 (x+y)+4`
` ⇒ 10y +x =6x + 6y + 4`
`⇒ 6x + 6y -10y -x=-4 `
` ⇒ 5x -5y =-4`
After interchanging the digits, the number becomes `10x + y.`.
If 18 is subtracted from the number, the digits are reversed. Thus, we have
` ( 10y + x )- 18 =10x + y`
`⇒ 10x + y -10y -x = -18 `
` ⇒ 9x -9y =-18`
` ⇒ x -y =-18/9`
` ⇒ x - y = -2`
So, we have the systems of equations
` 5x - 4y = -4 `
` x - y =-2`
Here x and y are unknowns. We have to solve the above systems of equations for xand y.
Multiplying the second equation by 5 and then subtracting from the first, we have
`(5x-4y)-(5x-5y)=-4-(-2xx5)`
` ⇒ 5 x -4y -5x +5y =-4+10`
` ⇒ y = 6`
Substituting the value of y in the second equation, we have
` x - 6=-2`
`⇒ x = 6-2 `
` ⇒ x =4`
Hence, the number is `10 xx6+4=64.`
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