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:



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



The quotient is 4 and the remainder is 3:



Solving for b, we see that b=7

This means a=4

The number is 47