maybe you have seen this before....

        FORTY 
      +   TEN 
      +   TEN 
      -------
      = SIXTY

If each letter represents a digit and all ten digits are to be included in the above equation. Can you find the arrangement of digits that would satisfy the equation? (TEN does not mean T x E x N. It represents the places the digits take)

First, we have Y + N + N = Y and T + E + E = T. That means N and E must be 0 and 5. If N was 5, there would be a 1 carried over, and there is no solution for E then. So, N=0 and E=5.

Second, we have FOR + T + T + carry = SIX. Since both of the top digits changed, that means there is a carry to the column with the O and the F. Since the 0 is already used, and the maximum carry is 2 (9+9+9 = 27 for the first column, 9+9+9+2=29 for any other column), that mean we must be adding enough to roll over from 9 to 1 in the 1000's column, a carry of 2 from the 100's column. So O=9, I = 1, and S = F+1. This also forces S to be 3,4,7, or 8, since it can't be either a used number or one more than a used number. Likewise, F must be 2,3,6, or 7.

Third, to have a carry of 2, that means R+T+T must be greater than 20. Since 0 and 1 are both used, that means x is at least 2, so R+T+T is between 22 and 28. If T was the highest unused value (8), and since we have a carry of 1 (T + 5 + 5 = T, carry 1), R would have to be 6 or 7 (5 is already used). Likewise, if R was 8, T would have to be 7. So R is 6, 7, or 8; and T is 7 or 8.

Now, two out of 6, 7, and 8 are used for R and T, so since S and F must be consecutive, they now must be 2, 3, or 4. Since either S or F will have be be 3, that raises the new minimum for R+T+T to be 24, forcing R to be 7, T to be 8, and X to be 4. This then forces F to be 2, and S to be 3.

Thus far, we have N=0, E=5, O=9, I=1, R=7, T=8, X=4, F=2, and S=3. The only usused letter is Y, and the only unused value is 6, so Y=6, giving us: 29786 + 850 + 850 = 31486, which checks.

Posted by Ender on 2002-07-08 10:45:43