An alphanumeric

Freddie Manners, of Packwood Haugh School in Shropshire, has e-mailed us about the problem "Cayley" from the February 2001 magazine. He has noticed that we had given more information than was needed to solve the problem. He is absolutely right: in fact when the problem was originally invented, it was:

Given that the letters in this sum represent the digits 1 ... 9, what number is represented by "CAYLEY"?

The people trying to solve it then were members of the Cambridge, Oxford and Warwick University maths societies, but they only had a limited amount of time in which to do it. We thought that maybe you would need some help, which is why we told you the numbers represented by A and D, but Freddie has proved us wrong!

Here is his reasoning:

1. D is even, as it comes from Y+Y.
2. If U is less than 5, column 3 will not carry, and so the U in column 2 will be even.
   
   If U is 5 or more, column 3 will carry, and so the U in column 3 will be odd.
   
   This means that U can only be 2, 4, 5, 7 or 9.
3. C must be less than 5, as column 1 does not carry.
   
   This also means that column 4 will not carry, and so C in column 3 must be even.
   
   This means that C must be 2 or 4.
4. If C is 2, then from column 3, U must be 1 or 6. But from (2) above, this is not possible.
   
   So C is 4.
5. Because C is 4, then either E=8 and L=9 or E=9 and L=8. Either way, this rules out any other letter being 8 or 9.
6. Looking at column 3, because C is 4, U must be 2 or 7. Both of these are OK by (2).
7. Try U=2. We now have U=2, C=4, and E/L=8. The only even number left is 6, so D is 6.
   
   From column 2, A is 1 or 6, so A must be 1.
   
   From column 6, Y is 3 or 8, so Y must be 3.
   
   We now have:

   4	1	2	4	H	3
   4	1	2	4	H	3
   -	-	-	-	-	-
   8	2	4	9	I	6

   
   
   H and I must be 5 and 7, which doesn't work. The only guess was that U was 2, so that must have been wrong.
8. We now know that U must be 7.
   
   From column 2, A is 3 or 8, and 8 is used, so A must be 3.
   
   The only even numbers for D to be are 2 and 6. But if D is 6, then Y must be 3 or 8, and they're both used up.
   
   So D is 2.
   
   Putting this in gives us:

   4	3	7	4	H	Y
   4	3	7	4	H	Y
   -	-	-	-	-	-
   8	7	4	9	I	2

   
   
   
9. We now have just H, I and Y left, which must be 1, 5 and 6. It is quick to check and see that the only possibility now is

   4	3	7	4	5	6
   4	3	7	4	5	6
   -	-	-	-	-	-
   8	7	4	9	1	2

   
   
   

Well done to Freddie, both for spotting that the hints were unnecessary, and for explaining so clearly why.