A two-digit number is such that the product of its digits is 6. If 9 is added…

For any two-digit number, if the tens digit is x and the units digit is y, the number itself equals 10x + y, while the number obtained by swapping its digits equals 10y + x. Any condition about the digit product, or about what happens when the digits are interchanged, becomes a plain algebraic equation once written this way.

1. Let the required number have tens digit x and units digit y, so the number is 10x + y.

2. The digit-product condition gives xy = 6.

3. Adding 9 reverses the digits: (10x + y) + 9 = 10y + x, which simplifies to 9 = 9(y − x), i.e. y − x = 1.

4. Substituting y = x + 1 into xy = 6 gives x(x + 1) = 6, i.e. x2 + x − 6 = 0, which factors as (x + 3)(x − 2) = 0.

5. Since x is a tens digit it must be a positive integer, so the only valid root is x = 2, giving y = 3. The number is 10(2) + 3 = 23.

Checking directly: 2 × 3 = 6, so the product condition holds, and 23 + 9 = 32, which is exactly 23 with its digits swapped. Both conditions hold only for this value, confirming the result independently of the algebra above.