Yes, the solution is unique...
The solution to the sum
THREE
THREE
ONE
-----
SEVEN
where distinct letters represent distinct digits and THREE is divisible by 3, SEVEN is divisible by 7 and there are no leading zeroes is
23577
23577
817
-----
47971
This is a unique solution and the proof (which took considerably longer to draft than to develop) follows.
Part 1: Solutions for E and N
From the rightmost column of the sum we have
N = 3*E (mod 10) [Equation 1]
c10 = (3*E - N)/10 [Equation 2]
and from the 2nd rightmost column we have
E= 2*E+N+c10 (mod 10) [Equation 3]
where c10 is the number of 10's carried in the sum from the rightmost column (the units) to the 2nd rightmost column
The 10 possible values for E are 0-9. For each possible value of E use Equations 1 and 2 to calculate corresponding values for N and c10 and then apply these in Equation 3 to calculate E. If the calculated value corresponds to the value applied in [1] and [2] then the resulting value of E is a possible solution in the sum. The 10 possibilities are evaluated below.
E N c10 E' Solution
---------------------------
0 0 0 0 No
1 3 0 5 No
2 6 0 0 No
3 9 0 5 No
4 2 1 1 No
5 5 1 6 No
6 8 1 1 No
7 1 2 7 Yes
8 4 2 2 No
9 7 2 7 No
E=0 is not a valid solution (despite the fact that the value emerging from Equation 3 is also 0) because E and N must be distinct, which is not the case when E=0.
The only possible solution for (E, N) is (7, 1).
Part 2: Feasible values for H
From the 4th rightmost column of the sum we have
E = 2*H + c1000 (mod 10)
where c1000 is the number of thousands carried in the sum from the 3rd rightmost column to the 4th.
Applying the value derived for E from part 1, this means
7 = 2*H + c1000 (mod 10)
and this implies c1000 must be odd.
In fact, c1000 must be 1 since the maximum carry in thousands when adding three 3-digit integers is 2. Hence
2*H + 1 (mod 10) = 7 [Equation 4]
There are 8 possibilities remaining for H but only two of these, 3 and 8, satisfy Equation 4 and these are the only feasible values for H.
Part 3: Feasible possibilities for R, O and V
Given that E=7 and N=1 (Part 1) the carry in hundreds from the second rightmost (ie tens) column of the sum to the third rightmost is c100 = 1.
Therefore, the 3rd rightmost column of the sum yields
V = 2*R + O + 1 (mod 10) [Equation 5]
There are 8 possibilities for R. These are 0 to 9 excluding 1 and 7. There are 7 possibilities for O and these are 1 to 9 excluding 1 and 7. The resulting 56 possibilities for V arising are shown below.
. | O
V| 2 3 4 5 6 8 9
---------+------------------------------------------------
R 0| 3 4 5 6 7 9 0
2| 7 8 9 0 1 3 4
3| 9 0 1 2 3 5 6
4| 1 2 3 4 5 7 8
5| 3 4 5 6 7 9 0
6| 5 6 7 8 9 1 2
8| 9 0 1 2 3 5 6
9| 1 2 3 4 5 7 8
These 56 possibilities can be reduced by noting the conditions that
- R, O and V must be distinct from each other
- V cannot be 1 or 7 (see Part 1)
- The carry value c1000 must be 1 (see Part 2)
- The R, O, V values cannot contain both 3 and 8 (since this conflicts with the requirement that H is 3 or 8, see part 2)
Applying these conditions reduces the number of possibilities to just 13 as shown below.
. | O
V| 2 3 4 5 6 8 9
---------+------------------------------------------------
R 0|
2| 0 4
3| 2 6
4| 2 5 8
5| 3 4 9
6| 5 8
8| 9
9|
In a list format these are
R O V
2 5 0
2 9 4
3 5 2
3 9 6
4 3 2
4 6 5
4 9 8
5 2 3
5 3 4
5 8 9
6 2 5
6 5 8
8 2 9
Part 4: Feasible possibilities for R, O, V, H and T and the solution
The list for R, O and Vis first used to create a list for R, O, V and H. H takes the values of either 3 or 8 (see Part 2). In the list of 13 possibilities above 5 contain the value 3 (so require H to be 8), 4 contain the value 8 (so require H to be 3) and 4 contain neither 3 nor 8 (allowing H to be either 3 or 8).
This results in 17 possibilities for R, O, V and H which are listed below.
R O V H
2 5 0 3
2 5 0 8
2 9 4 3
2 9 4 8
3 5 2 8
3 9 6 8
4 3 2 8
4 6 5 3
4 6 5 8
4 9 8 3
5 2 3 8
5 3 4 8
5 8 9 3
6 2 5 3
6 2 5 8
6 5 8 3
8 2 9 3
From inspection of the sum, T cannot take values greater than 4 and cannot be 0 or 1. If H is 3, T cannot be 3 and if H is 8 there is a carry value into the leftmost column (c10000 in the notation adopted) of 1. In this case also T cannot be 3 because this would require S to be 7 in conflict with E.
Therefore T can only take values of 2 or 4. In the list of 17, above 7 rows contain the value 2 but not 4 and therefore require T to be 4. 4 rows contain the value 4 but not 2 and therefor require T to be 2. 3 rows contain both a 4 and a 2 and so are infeasible as an overall solution while the remaining 3 rows contain neither a 2 nor a 4 and so can be have a value of either 2 or 4 for T. Taken together, this yields 17 possibilities for R, O, V, T and H and these are listed below.
These possibilities along with the earlier result in Part 1 can be used to generate a value for THREE and this can be checked to determine if it is divisible by 3.
R O V H T THREE Div by 3
2 5 0 3 4 43277 No
2 5 0 8 4 48277 No
3 5 2 8 4 48377 No
3 9 6 8 2 28377 Yes
3 9 6 8 4 48377 No
4 6 5 3 2 23477 No
4 6 5 8 2 28477 No
4 9 8 3 2 23477 No
5 2 3 8 4 48577 No
5 3 4 8 2 28577 No
5 8 9 3 2 23577 Yes
5 8 9 3 4 43577 No
6 2 5 3 4 43677 Yes
6 2 5 8 4 48677 No
6 5 8 3 2 23677 No
6 5 8 3 4 43677 Yes
8 2 9 3 4 43877 No
The 17 possibilities for R, O, V, H and T are reduced to just 4 when divisibility of THREE by 3 is considered.
These 4 possibilities are now used to generate SEVEN (as the sum of 2*THREE + ONE) and to test its divisibility by 7.
R O V H T SEVEN Div by 7
3 9 6 8 2 57671 No
5 8 9 3 2 47971 Yes
6 2 5 3 4 87571 No
6 5 8 3 4 87871 Yes
Although 2 rows pass the divisibility by 7 test, the 4th row requires S and V to have a common value of 8 therefore this is not a solution to the sum.
This leaves the sole solution to the sum as:
23577
23577
817
-----
47971
modellingman