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प्रश्न
A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number.
उत्तर
Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is `10y + x`.
The product of the two digits of the number is 20. Thus, we have ` xy =20`
After interchanging the digits, the number becomes `10x + y`.
If 9 is added to the number, the digits interchange their places. Thus, we have
`(10y+x)+9=10x+y`
`⇒ 10y +x +9 =10x +y`
` ⇒ 1=x + y -10y -x =9`
` ⇒ 9x -9y =9`
`⇒ 9( x - y)=9`
` ⇒ x -y 9/9`
` ⇒ x - y =1`
So, we have the systems of equations
`xy=20`
`x-y=1`
Here x and y are unknowns. We have to solve the above systems of equations for xand y.
Substituting `x =1+y` from the second equation to the first equation, we get
`(1+y )y=20`
`⇒ y + y^2 = 20`
`⇒ y^2 + y = 20 =0`
` ⇒ y^2 + 5y -4y -20 =0`
`⇒ y (y + 5)-4(4+5)=0`
`⇒ ( y + 5)(y - 4)=0`
`⇒ y = -5` Or `y = 4`
Substituting the value of y in the second equation, we have
Hence, the number is `10 xx4 + 5= 45`.
Note that in the first pair of solution the values of x and y are both negative. But, the digits of the number can’t be negative. So, we must remove this pair.
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