Alphametic

[cinelli]

THREE
THREE
  ONE +
-----
SEVEN

In this alphametic each letter stands for a different digit. Leading zeros are not allowed. In addition THREE is a multiple of 3 and SEVEN is a multiple of 7.

Cinelli

[GoSeigen]

Spoiler:

THREE
THREE
  ONE +
-----
SEVEN

23577
23577
  817 +
-----
47971

Rough summary of method:
1. Starting with tens and units columns, the only combination which fits is E=7 and N=1. This leaves a carry of one for the hundreds.
2. The total of the thousands column is now known to be 7. Since H+H is even then there must also be a carry of one in the thousands, and H∈{3,8}
3. Trying H=3, then T∈{2,4} because the total is less than 100,000 and both 1 and 3 have been used.
4. With T=2, H=3 then R must 5 or 8 (2 is already taken) to make THREE a multiple of 3
5. Trying R=5 and evaluating while ensuring that the total is divisible by seven gives the solution above.

Slightly lucky that the first choice worked at each decision stage. I couldn't see any other obvious shortcuts but it would not take to long to work through the alternatives.


Neat puzzle.

G
P.S. I also didn't bother to check whether the solution is unique. I think working through the alternative values for H,T and R to find other solutions (if any) would be sufficient.

[9873210]

THREE
THREE
  ONE +
-----
SEVEN

In this alphametic each letter stands for a different digit. Leading zeros are not allowed. In addition THREE is a multiple of 3 and SEVEN is a multiple of 7.

Cinelli

You missed the opportunity to say that ONE is a multiple of 1 with a straight face.

23577
23577
  817 +
-----
47971

I have not checked that the solution is unique.
Notice that E forces N, and trying all 9 digits for E and looking at the units and hundreds only E=7 works.
This leaves six possible pairs for T and H. Two of them fall out because of duplicates, work through the other four until you find a solution.

[modellingman]

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

[cinelli]

Three fine solutions, the first two of which almost dead-heated, only two minutes apart. Excellent analysis from modellingman. Well done all.

Cinelli