Question

The number of two digits is such that the product of their digits is eighteen. When the numbers are subtracted 63, the digits of the numbers are reversed. Find the number.

Answer 1

Assumption

Two digit number be 10p + n

$\large{\fbox{Situation}}$

p × n = 18..........(1)

10p + n - 63 = 10n + p

9p - 9n - 63 = 0

9p - 9n = 63

p - n = 7..........(2)

p = 7 + n

$\large{\fbox{Substitute\:value\:of\:p\:in\:(1)}}$

( 7 + n ) × n = 18

n² + 7n = 18

n² + 7n - 18 = 0

n² + 9n - 2n - 18 = 0

n(n + 9) - 2(n + 9) = 0

(n + 9)(n - 2) = 0

n = -9

n = 2

$\fbox{Negative\;value\;is\;not\;acceptable}$

Hence,

n = 2

p = 7 + 2

p = 9

$\large{\fbox{Two\;digit\;number}}$

= 10(9) + 2

= 92

Answer 2

$\huge\bf{Answer:-}$

Let p be the product and n be the number.

Product × Number = 18 Equation - (1)

$10 + n - 63 = 10n + p$

$9p - 9n - 63 = 0$

$9p - 9n = 63$

$p - n = 7 Equation - (2)$

$p = 7 + n$

Adding values of p for Equation - (1)

$7 + n × n = 18$

n² + 7n = 18

n² + 7n - 18 = 0

n² + 9n - 2n - 18 = 0

$n*n + 9 - 2*n + 9 = 0$

$n + 9*n - 2 = 0$

$n = -9$

$n = 2$

This negative values is not correct so,

2 = number

7 + 2 = 9 is the product

Therefore, product = 9

The Number =

$= 10*9 + 2 = 92$

Therefore, 92 is the two digit number