Assign the digits 0 through 9 to the letters below to create valid sums. Each letter stands for a unique digit, and all occurrences of that letter stand for the same digit. (For instance, if A = 6, then all A’s in both sums are 6’s, and no other letter stands for 6.) The words represent three- or four-digit numbers. (“ALL” is a three-digit number whose latter two digits are the same.) Leading zeros are not allowed—098 is not a valid representation of the number 98.
You don’t need extensive trial and error. Recall that when adding two numbers, every column either produces no carryover to the next column or produces a carryover of 1. The first deducible letters are C, R and A. After that, make a table to track the possible values for some of the remaining letters.
Because it involves many variables and cases, solutions to this type of problem can be cumbersome to read when written out. Still, we’ve included a step-by-step solution below for anybody who wants to track through the reasoning.
C = 1 because two three-digit numbers (ALL and ACE) cannot sum to a number CREW in the two thousands or higher (999 + 999 = 1,998). In the second-to-last column, we have R in the same position in all three words. This can only happen in two ways: either R = 0, and D + D did not produce a carryover, or R = 9, and D + D did produce a carryover (9 + 9 + 1 = 19, which would place a 9 in the R spot in GURU). R cannot be 9, though, because A + A cannot produce a 9 in the R location of CREW. The only way it could do so while producing the necessary carryover for C would be if A = 9, which isn’t allowed because we’ve already assigned 9 to R in this scenario. So R = 0. Now A = 5 because that’s the only way for A + A to produce a 0 in the R location of CREW.
We know from earlier that D + D does not produce a carryover, meaning that D = 2, 3, or 4. (The digits 0 and 1 have already been assigned, and anything larger than 4 produces a carryover.) Each of these three possibilities would immediately force the values of U, O and G, as summarized in the following table:
For example, if D = 2, then U = 4 (because U = D + D). Then, in the second column of the second sum, we have A + O producing the first U in GURU. We know A = 5, and in this hypothetical, U = 4. The only value for O that makes this work is O = 9 because 5 + 9 = 14. This would place a 4 in the correct position of GURU and produce a carryover into the C + C column, making G = 3. (Recall C = 1.) A similar chain of reasoning yields the other two columns of the table.
We can immediately strike the middle column from our possibilities because 1 has already been assigned to C. In the remaining two possible scenarios, notice that 2, 3 and 4 are assigned in both of them, along with either an 8 or a 9. So the only unassigned numbers left for L, E and W are 6, 7, 8 and 9. The last column of the first sum only works if 7 and 9 are added to yield 16, making W = 6. (It can’t be 8 + 9 because we need one of those to fulfill the letters in the table.) The L + C column in the first sum only works if L = 7 and E = 9. (Remember C = 1.) With 9 assigned, only the third column in the table remains possible: D = 4, U = 8, O = 3 and G = 2.
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