Question

The product of digits of a two digit number n is m . Find n if m+n=118.

Answer 1

$$
n=a10+b\\
a\times b=m\\
m+n=118\\
m=118-n\\
a\times b=118-n\\
b=\cfrac{118-n}{a}\\
a=\cfrac{n-b}{10}\\
b=\cfrac{10(118-n)}{n-b}\\
nb-b^2=10(118-n)\\
b^2-nb+10(118-n)=0\\
$$
b can be solved using the quadratic formula in terms of n, from which a can be solved in terms of n.
This can then be used with the following to find n:
$$
a\times b=118-n\\
$$

Answer 2

There are 16 solutions. We have 3 equations but 4 unknowns:

http://www.wolframalpha.com/input/?i=n%3Da*10%2Bb%2Ca*b%3Dm%2Cm%2Bn%3D118

Answer 3

Another way to solve this is:
\[n=10a+b \rightarrow m=ab \rightarrow m+n=10a+b+ab=10a+b(a+1)=118\]

Now a is a number from 1...9 and b is from 0..9 - since n is given to be a 2 digit number

Notice that if a=5 then the maximum value of \[10a+b(a+1)\] can be 10*5+9*6=104.
So we know that a has to be 6...9.

By simple trial and error we can find: a=7 and b=6 thus giving n=76

Checking out: 7*6+76=42+76=118

Here, using the fact that 1<=a<=9 and 0<=b<=9 makes all the difference and avoids the use of quadratics and the like.

Answer 4

Thanks! @ybarrap @RajeshRathod

Answer 5

yw - note that a and b can either be both negative or both positive since the sign of n was not given.