Question

A two digit no.is such that Product of its digits is 18 when 63 is added to the number the digits interchanges Thier place. Find the number

Answer 1

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Answer 2

Hello there!
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Let's say that the two digit number, "n" is written in the form "XY"

But mathematically, it would be written as:

10x + y

Where X is the ten's digit number and Y is the unit's digit number.

Hence by we have been given in the question that:

$x \times y = 18$
Also that when 63 is added to n the digits get reversed.

Hence the reversed number would be

10y + x

Hence we would write the equation as :

$10x + y + 63 = 10y + x$
Now we will solve this equation.

$10x + y + 63 = 10y + x \\ 9x + 63 = 9y \\ x + 7 = y$
Remember that both X and Y are digits hence they can only be integers from 0 to 9.

$xy = 18 \\ x = \frac{18}{y}$
Now we substitute this value of X in the original equation.

$\frac{18}{y} + 7 = y \\ 18 + 7y = {y}^{2} \\ {y}^{2} - 7y - 18 = 0 \\ {y}^{2} - 9y + 2y - 18 = 0 \\ y(y - 9) + 2(y - 9) = 0 \\ (y + 2)(y - 9) = 0$
Hence we can infer from the above equation, that Y is -2 or 9. But since we know that Y is a digit as I explained before, we know for sure that Y = 9

Hence we can now find x:

$x = \frac{18}{y} \\ x = \frac{18}{9} \\ x = 2$
Hence n, the original number is

$n = 10x + y \\ n = 10 \times 2 + 9 \\ n = 20 + 9 \\ n = 29$
Hence the number is 29.
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Hope this helps!