Finding the two digit number. A two digit number is such that the product of its digit is 18.When 63 is subtracted from the number , the digits interchange their places.Find the number. 
 A: Alternative solution:
18 may be factorised into products of single digit numbers as 2*9 or 6*3
We know the number is greater than 63, so first digit must be greater than 6.
Only one possibility.
Explanation:
Algebra is very often a sledgehammer in cases like this. The algebraic answer will always work, and is valid whether numbers are integers, reals, or complex (or...) This alternative is just a shortcut which exploits the fact that we are dealing with integers (a 
UFD) which have unique factorisations, but not real numbers (a 
field) in which every number is a factor of every other number (except zero). 
Because discrete integers and continuous real numbers are quite different, they can have very different properties. So problems which are hard in one domain can be easy in another. For instance finding the log of a number is easy in real numbers, but can be nearly impossible in discrete maths, which is the basis of modern cryptography. 
A: 
A two digit number

Let $n = 10a+b$. 

is such that the product of its [digits] is 18.

$ab = 18$

When 63 is subtracted from the number, the digits interchange their places.

$10a+b-63 = 10b+a$

Find the number.

$n =\;?$ 
Can you see where to go from here?
A: Let the number be n and x and y be tens and unit digit 
so 
n=10x+y
now, x*y=18
Also, n-63=10y+x from condition 2
then,
10x+y-63=10y+x
or 9x-9y=63
x-y=7
possible values of x and y - 8,1 & 9,2
which gives us only 9,2 as the possible answer
so the number is 92.
If you want to solve using quadratic equations then substitute y by 18/x and solve.
Check:
92-63=29 (numbers interchanged places)
PS: I just joined this forum today, so I am not an expert of formatting. :)
A: Let the number be=xy
x*y=18
63-xy=yx
xy may be=
6*3=18
9*2=18
63-63=0
92-63=29
Therefore the number is 92
