Two and Two

There were seven solutions to this alphanumeric, and four people or groups found all seven: Andrei (School 205, Bucharest, Romania), Sim (Raffles Girls' Primary, Singapore), Prateek (Riccarton High School, New Zealand) and Damen, Katrina and Oliver (Crownfield Junior School, England). A number of pupils from Crownfield Junior School produced good work on this problem.

The key to this sort of problem is to be systematic, and Andrei explained very clearly how he worked through every possibility. Here is his solution:

I started from the observation that I must add two 3-digit numbers, and the result is 4-digit one, so that I must assign the value 1 to the letter F.

My method is to give values to the letters from right to left, the way additions are performed. [O is also a good place to start as it occurs three times. - Ed.]

• If O=0, then R must be 0 too. As each letter stands for a different number, it isn't possible.
• O can't be 1, as F=1.
• If O=2, then

  		T	W	2
  +		T	W	2
  -	-	-	-	-
  	1	2	U	R

  This means R=4 and T=6. In this case W+W=U, and I analyse the numbers for W one by one:
  • W=0 NO, because 0+0=0, so U must be 0 too.
  • W=1 NO, because F=1.
  • W=2 NO, because O=2.
  • W=3, so U=6. NO, because T=6.
  • W=4 NO, because R=4.
  • W=5 NO, because this would make U greater than 9.
• If O=3, then

  		T	W	3
  +		T	W	3
  -	-	-	-	-
  	1	3	U	R

  There is no solution, because both R and T must correspond to 6.
• If O=4, then

  		T	W	4
  +		T	W	4
  -	-	-	-	-
  	1	4	U	R

  This means R=8 and T=7. In this case, W+W=U, and analysing as previously, I find that W cannot be 0, 1, 2, 4, and cannot be greater than or equal to 5. There is a solution with W=3 and U=6.
• If O=5, then

  		T	W	5
  +		T	W	5
  -	-	-	-	-
  	1	5	U	R

  This means T=7 and R=0. Now, W must be at least 5, because W+W+1=10+U.
  • W=5 NO, because O=5.
  • W=6, U=3 - SOLUTION
  • W=7 NO, because T=7.
  • W=8, U=7 NO, because T=7.
• If O=6, then

  		T	W	6
  +		T	W	6
  -	-	-	-	-
  	1	6	U	R

  This means R=2 and T=8. W+W+1=U, so W=4.
  • W=0, so U=1. NO, because F=1.
  • W=1 NO, because F=1.
  • W=2 NO, because R=2.
  • W=3, U=7 - SOLUTION
  • W=4, U=9 - SOLUTION
• If O=7, then

  		T	W	7
  +		T	W	7
  -	-	-	-	-
  	1	7	U	R

  This means R=4 and T=8. W+W+1=10+U. So, W=5.
  • W=5, so U=1. NO, because F=1.
  • W=6, so U=3 - SOLUTION
  • W=7 NO, because O=7.
  • W=8 NO, because T=8.
  • W=9, so U=9 - NO
• If O=8, then

  		T	W	8
  +		T	W	8
  -	-	-	-	-
  	1	8	U	R

  This means R=6 and T=9. W+W+1=U. So, W=4.
  • W=0, so U=1. NO, because F=1.
  • W=1 NO, because F=1.
  • W=2, U=5 - SOLUTION
  • W=3, U=7 - SOLUTION
  • W=4, so U=9. NO, becuase T=9.
• If O=9, then

  		T	W	9
  +		T	W	9
  -	-	-	-	-
  	1	9	U	R

  This means R=8 and T=9, but this gives no solutions because O=9.

So, the only solutions are: