Several of you sent correct or part answers so well done but they were not solutions. I have been very strict with you. If you have not given an explanation for your answer I have not credited you with a solution. The aim is not simply to find an answer but to explain to others how you arrived at it. This is an important part of mathematics that we are encouraging you to be involved in.

The two solutions I have used to form the basis of the one given below are from Clement Goh (River Valley High School, Singapore) and Andrei Lazanu (School 205, Bucharest). Well done both of you and thank you.

In order to understand it more easily, I changed the question to:

This is actually the same as the question that was given:

I just rephrased the question a little but it is still the same.

So you look at the changed one to know what I am talking about.

From the first column, we know that:
F=0

But we cannot say anything about R as we can't refer to anything to give us clues.

Either L+A = E or L+A = 10+E

If L+A = E
Then from the second column of the calculation
E+L = A (a)
This would only work if L = 0 but we already know that F=0
So L+A=10+E (b)

Therefore, substituting for A from (a) in (b):

L+E+L = 10+E
2L =10
L=5

A must be more than 6 to make 5+A more than 10.

So the possible solutions are:

A cannot be 7 or 9 because there will be two 5s, and two 9s respectively representing two different letters.