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Alphametic

cinelli
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Alphametic

#667047

Postby cinelli » June 2nd, 2024, 10:05 am

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
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Re: Alphametic

#667099

Postby GoSeigen » June 2nd, 2024, 4:40 pm

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.
Last edited by GoSeigen on June 2nd, 2024, 4:45 pm, edited 1 time in total.

9873210
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Re: Alphametic

#667100

Postby 9873210 » June 2nd, 2024, 4:42 pm

cinelli wrote:
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
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Re: Alphametic

#667110

Postby modellingman » June 2nd, 2024, 6:23 pm

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
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Re: Alphametic

#667489

Postby cinelli » June 4th, 2024, 7:26 pm

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

Cinelli


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