Brenda's younger brother
+ M A F O L A T F T H R H E E E R D R
Every letter in the above cryptarithm represents a distinct single-digit positive integer except for L which represents 0. Compute the 6-digit integer A L F R E D .
The answer is 802794.
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2 solutions
shit . genius as hell .
Another choice is [3 0 1 7 9 4], which is A=3, M=2, F=1, O=6, T=5, H=8, E=9, R=7, D=4. Try it.
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I think that isn't possible. If T = 5, T+T+1 = 11 mod 10 = 1 that generate carry over. (OK) If A = 3 and O = 6, A+O+1 mod 10 = 10 mod 10 = 0 = L (OK) but this generate also carry over. If F = 1 (OK) and M = 2, with A = 3. F+M+1 mod 10 = 4 (WRONG) :)
From the column (2) we know that E+E = E. This is possible only admit that R+R > 10. In particular it say (2R+1) mod 10 = R. So this is possible only if E = 9. So it means that R = {6, 7, 8}. From the column (3) we know that 2H + 1 mod 10 = R (because E = 9, and this is sure). The two value that respect the condition are 8 and 3. I will try with for absurd with one of the two, and if i don't conclude anything I demonstrate that is the other. So H = 3 means R = 7 and D = 4 So the set that we construct is {E, H, R, D} = {9, 3, 7, 4}. From the column (4) we know that T+T mod 10 = F. So this sum generate carry over. Which the remained values {1, 2, 5, 6, 8} this is possible only in three case: 1+1 mod 10 = 2 6+6 mod 10 = 2 8+8 mod 10 = 6 So it means that {T, F} = {6, 2}, {8, 6}, {1, 2} From the column (5) we know that A+O mod 10 = 0. But with the remained value this is possible only in two situation: {A, O} = {8, 2}, {8, 1} This two case exclude that {T, F} could be {8, 6} The last column say that F+M + 1 = A. But A could be only 8. So F+M = 7. We have that the possible combination of the set is {A, O} intersect {T, F} = {A, O, T, F} = {8, 6, 1, 2} With the remained value this is possible only if F = 2 and M = 5 So the complete construction is {A, D, E, F, H, L, M, O, R, T} = {8, 4, 9, 2, 3, 0, 5, 6, 7, 1} and so ALFRED = 802794.
I'm Batman addicted.
E - 9
R - 7
D - 4
H - 8
T - 5
F - 1
A - 6
O -3
M - 4
165897 + 435897 = 601794
The easiest would be to look at the second right column first. Either 2E=0 or 1+2E=E (mod 10). The first case is out since L is the only letter equal to 0. Hence we have 1+2E= E (mod 10) which means 1+2E=10+E or E=9. This means not only must there be carry over from the rightmost column, there is also carry over affecting the third right column since 1+2(9) = 19.
In other words, R+R=2R must be greater than 10 (10 is excluded since D>0). Since 9 is taken by E, R is at most 8. 2R greater than 10 means R is greater than 5 or R is at least 6. Hence R=6 or 7 or 8. Now observe that with the carry over affecting the third right column, 1+2H=R (mod 10). This means R must also be an odd number so the only possible digit for R would be 7. D=2R = 14 = 4 (mod 10) so D=4.
Now note that since 1+2H=7 (mod 10), H=3 or 8. But if H=8, the unused digits would be 1, 2, 3, 5 and 6. Observe the second left column which suggests that A+O=L. Since L=0, A+O=9 (with carry over) or 10. The only possibility for this is 3 and 6. But with carry over from the third left column, we must have 1+2T= 10+F. The only possible combination for this is F= 3 and T=6 again, resulting in duplications. Hence we must have H=3.
The unused digits would then be 1, 2, 5, 6 and 8. There is no carry over affecting the third left column so 2T= F (mod 10). We have (T, F)=(1, 2), (6, 2) and (8, 6). For the first casez there is no carry over affecting the second left column so A+O =10. We are left with 5, 6 and 8 which give no possible combinations. For the second and third cases, there is carry over affecting the second leftncolumnnso A+O=9. For the third case, we are left with 1, 2 and 5 which give no possible combinations again So we only need to check the second case. We are left with 1, 5 and 8. Now 1+8=9.
We have T=6 and F=2. We need A=8 and O=1 instead of the other way around since F+M=A. Now note that the leftmost column is affected by carry over so 1+2+M=8 which means M=5. This digit has not been used earlier so it's fine. Our answer is 802794.