Home Up Leave Message

Let's start with a basic problem:

 ONE
 ONE
 ONE
+ONE
----
 TEN

The idea is to substitute all the alphabetic characters with unique digits and have a valid mathematical statement. Thus we have each character from 0 to 9, but O may not be zero and none of O,N,E or T must be the same. We need to remember that there could be "carry overs" when adding the digits above the line.

We can now write above in another way:

(1)	E*4 = N + 10*b	(Where b is the "carry over")
(2)	N*4 + b = E + 10*a	(Where a is the "carry over")
(3)	O*4 + a = T	(Here is no "carry over")

Algebraic approach

We first need to take into account that no "carry over" can be more than 3. (Max digit would be 9 and 9*4 = 36 which means a "carry over" of 3) We know that 1 < T < 10 and 1 < O. But O must be less than 3, seeing that 3*4 = 12 and we are restricted to T < 10. Now we have O to be either 1 or 2.
We can rewrite (1) above and then combine it with (2):

(4)	N = E*4 - 10*b	(Swapped left and right side and take 10*b over)
(5)	E*16 - 40*b + b = E + 10*a	(Replace N*4 in (2) with (4)*4)
(6)	E*15 = 39*b + 10*a	(Take Es to left side and all a and b elements to right side)

If b is taken to be zero (no "carry over" when the Es are added together) then E*15 = 10*a. If E is zero then a must be zero but then N is also zero which breaks our original premises. The next possible solutions is when a=3, then E*15 = 10*3 or E = 30/15 = 2.

With E=2 we get N=E*4=8 (with no "carry over"). Now we can apply the values for E and N to (2) and we get 8*4 + 0 = 2 + 10*a or 32 = 2 + 10*a which we can write as a = 30/10 = 3, which we already determined (this just confirms it).

Now we substitute a into (3) and get O*4 + 3 = T. We know from above that O can only be 1 or 2, but if O is 2 then T = 2*4+3 = 11 which won't work. Thus O can only be 1 and this leads to 1*4 + 3 = 7 = T.

Now we have enough to rewrite the original problem with digits:

 182
 182
 182
+182
----
 728

Search Approach

With this approach we will go through all combinations of digits for O,E,N and T until we get values that work. N and T can be determined once we have values for O and E. Thus we only need to cover 9*9 = 81 combinations to search for an answer (remember that O and E must not be the same).

From above we know N = E*4 - 10*b also N = (E*4) mod 10, the carry = (E*4) div 10
Thus E = (carry + N*4) mod 10 and the new carry = (carry + N*4) div 10. This E must equal the original E.
Now T = O*4 + carry where we know O can be only 1 or 2

O	E	N	E2	T
1	0	0	0	0	N=2, so it's a no go
1	2	8	2	7	Got an answer

For this simple example the "Search Approach" will definately be the fastest approach. When multiple variables need to be searched then searching can be time intensive.

Programmatic Approach

A basic algorithm based on the Search Approach with some properties determined with the Algebraic Approach can help us to approach the problem with a program.

for O = 1 to 2
    for E = 0 to 9
        if (O <> E) then
            tmp = E*4            
            carry = tmp div 10
            N = tmp mod 10
            tmp = carry + N*4
            carry = tmp div 10
            E2 = tmp mod 10
            T = carry + O*4
            if (N <> O) AND (N <> E) AND (E2 = E) AND (T < 10) AND (T <> O) AND (T <> E) AND (T <> N) then
                print the results
                stop the program
            endif
        else
            break;
        endif
    endfor
endfor

Program Source