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Question
A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number.
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Solution
Let the digit at the tens place be ‘a’ and at units place be ‘b’.
The two-digit so formed will be 10a + b.
According to the first condition, the product of its digits is 6.
⇒ a x b =6
`=> x = 6/b` ...(1)
According to second condition
10a + b + 9 = 10b + a
`⇒ 9a - 9b = 9
`=> a - b = 1`
`=> a - 6/a= 1` From 1
`=> a^2 - a - 6 = 0`
`=> (a - 3)(a + 2) = 0`
`=> a= -3 or 2`
Since a digit cannot be negative, a = 2.
`=> b = 6/a = 6/2 = 3`
Thus, the required number = 10a + b = 10(2) + 3 = 23
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