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Question
A two digit number is such that the product of the digits is 12. When 36 is added to this number the digits interchange their places. Determine the number.
Solution
Let a digit at unit's place be x and at ten's place by y.
then according to problem
Required no.
= 10y + x
On interchanging the digits
Number formed
= 10 x + y
xy = 12
∴ x = `(12)/y`
10y + x + 36 = 10x + y
10y + x - 10x - y = -36
9y - 9x = -36
9(y - x) = -36
y - x = `-(36)/(9)`
y - x = -4
On substituting value of x = `(12)/y`
`y - (12)/y` = -4
`(y^2 - 12)/y` = -4
y2 + 4y - 12 = 0
y2 + 6y - 2y - 12 = 0
y (y + 6) -2 (y + 6) = 0
(y + 6) (y - 2) = 0
y = -6, 2
When y = 2
x = `(12)/(2)` = 6
Required no.
= 10y + x
= 10 x 2 + 6
= 20 + 6
= 26.
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