The tens digit of a number is 3 less than the units digit. If the number is divided by the sum of the digits, the quotient is 4 and the remainder is 3. What is the original number?

You could probably work this with trial and error, but that's no fun.

The tens digit is 3 less than the units digit:

Let a=tens digit and b=units digit.

a=b-3

The number is divided by the sum of the digits:

We can represent the number by

10a+b

and when dividing by the sum of the digits we get:

\frac{10a+b}{a+b}

Make the sub a=b-3 into this and we get:

\frac{11b-30}{2b-3}

The quotient is 4 and the remainder is 3:

\frac{11b-30}{2b-3}=4+\frac{3}{2b-3}

Solving for b, we see that b=7

This means a=4

The number is 47