First, since there are no zeroes and all digits are distinct, O + S cannot be 2. Thus, O + S is either 11 or 12, but there is a carry digit in either case. So, N + A = 6.
Similarly, E + A cannot be 1; either E + A = 10 or E + A = 11.
Suppose E + A = 11. Then, R + L = 3 (with no carry digit), so R and L are 1 and 2 in some order. Also, I + R = 5 (there is a carried 1 into that column, and I cannot be big enough to make the column total 15). Additionally, if E + A = 11, and N + A = 6, then N + 5 = E. So, N cannot be 5 or 6 (that would make E 10 or 11). Also, N is not 1 or 2 (R and L are). If N = 3, then A = 3 (N + A = 6), and if N = 4, A = 2, but again either L or R is 2. Therefore, the assumption that E + A = 11 must be false.
So, E + A = 10, and there is a carry digit from the ones column, so R + L = 13.
Since N + A = 6 and E + A = 10, N + 4 = E. So, N cannot be 6 (E would be 10). If N = 3, then A = 3, which is not possible; N (or A) cannot be 3). If N = 1, then A = 5, but if N = 1, E = 5, which is also not possible.
So, there are three valid possibilities so far for N, A, and E:
N = 2, A = 4, E = 6;
N = 4, A = 2, E = 8; or
N = 5, A = 1, E = 9.
Since R + L = 13, R and L must be (in some order) 4 and 9, 5 and 8, or 6 and 7. However, each of the possibilities for N, A, E includes a 4 or a 9, so R and L must be 5 and 8 or 6 and 7.
Therefore, the smallest value R can take is 5, but since there is a carry digit into the I + R = 6 column, and I cannot be zero, I + R must equal 15 (not 5). R cannot be 5 (I would have to be 10).
Since R + I = 15 and R + L = 13, L + 2 = I.
Therefore, there are three sets of possible values for R, L, and I:
R = 6, L = 7, I = 9;
R = 7, L = 6, I = 8; or
R = 8, L = 5, I = 7.
Comparing the sets of values for N, A, E with the possibilities for R, L, I, we can pair them up.
If E = 6, R or L cannot be six, so the first set for N, A, E is only possible with the third group of R, L, I.
Similarly, if E = 8, neither I nor R can be 8, so the second N, A, E is only possible with the first R, L, I.
Finally, if N = 5, L cannot be 5, and if E = 9, I is not 9, so the third N, A, E matches the second R, L, I.
There are three possibilities, namely:
N = 2, A = 4, E = 6, R = 8, L = 5, I = 7;
N = 4, A = 2, E = 8, R = 6, L = 7. I = 9; or
N = 5, A = 1, E = 9, R = 7, L = 6, I = 8.
There are three variables left to look at (S, T, O), and two possibilities for the sums of the remaining columns. Either S + T = 4, O + S = 12, and T + 8 = O, or S + T = 14, O + S = 11, and O + 3 = T.
If, in the first case, T + 8 = O, the only possible values are T = 1 and O = 9, with S = 3.
We need to look at the above three cases for the other six letters to see if there is one in which those values have not yet been assigned, and we find that the first set fits.
So, one possible set of values is:
N = 2, A = 4, E = 6, R = 8, L = 5, I = 7, T = 1, O = 9, S = 3.
Given these values, 725613 becomes INLETS, which is indeed a word.
Just to be thorough, check the second case above, wherein S + T = 14, O + S = 11, and O + 3 = T. The greatest O can be is 6, but since 6 is already represented in each of the above possibilities for the other six letters, O can't be six. If O is 5, T is 8, but all of the choices above have a letter for 5 or 8 already as well. Similarly, 4 and 7, 3 and 6, or 2 and 5 are not possible either. If O = 1 and T = 4, S would have to be 10 (to make O + S = 11); so we see that there are no remaining possibilities for the value of O.
Thus, the unique solution is as follows:
T = 1
N = 2
S = 3
A = 4
L = 5
E = 6
I = 7
R = 8
O = 9
and the word in question is INLETS.
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Posted by DJ
on 2003-12-28 17:39:42 |
No Subject
