Let the Lass...

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Posers and Puzzles 29 Jun '20 17:05
  1. R
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    29 Jun '20 17:053 edits
    Have a go at the following cryptogram!

    ......L E T
    +...T H E
    ____________
    …L A S S

    Each of the Letters represents a distinct digit 0 - 9 and none of the numbers can lead with 0 ( i.e 053 = "THE" is not valid ).

    There are multiple solutions, what are the letters ( digits ) that are common to all of them?
  2. Subscribervenda
    Dave
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    29 Jun '20 19:30
    @joe-shmo said
    Have a go at the following cryptogram!

    ......L E T
    +...T H E
    ____________
    …L A S S

    Each of the Letters represents a distinct digit 0 - 9 and none of the numbers can lead with 0 ( i.e 053 = "THE" is not valid ).

    There are multiple solutions, what are the letters ( digits ) that are common to all of them?
    The numbers are 148 and 874
    L=1
    E=4
    T=8
    H=7
    A=0
  3. R
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    29 Jun '20 19:471 edit
    @venda said
    The numbers are 148 and 874
    L=1
    E=4
    T=8
    H=7
    A=0
    You have found 1 of the multiple solutions, congrats. However, all of solutions are required to answer the question.
  4. Standard memberBigDogg
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    30 Jun '20 01:412 edits
    Reasoning:
    I. L<>0 and T<>0 [both are leading digits]
    II. L = 1, because it is a carried digit, and the max (and only) carried digit possible is 1 [9 + 9 = 18].
    III. T + E = LS [column 1; (T + E) must be > 9, since T + a different digit also sums to S, and the only way it can do that is with a carried L]
    IV. H + 1 = T [column 2, given that it receives a carried L]
    V. 2 + T = LA [column 3]

    Plugged in the lowest possible T, T = 8.
    Now, by V., A = 0.
    By IV., H = 7.
    By III., and avoiding repeated digits, 6 >= E >= 4 and 2 <= S <= 4.

    Next, tried T = 9.
    By V., A = 1 --- but L already = 1, so this is not distinct!

    Therefore, in ALL solutions:
    T = 8; A = 0; H = 7; L = 1
  5. Subscribervenda
    Dave
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    30 Jun '20 08:09
    @joe-shmo said
    You have found 1 of the multiple solutions, congrats. However, all of solutions are required to answer the question.
    Do we get to know how many possible solutions there are?.
    Bigdogg's logic is sound but I worked it out less logically.
    L has got to be 1 because the total cannot be greater than 1,000
    S therefor cannot be 1 but could be 2,so I tried to make the sum fit by trying the lowest possible total i.e 1022 and so on
  6. R
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    30 Jun '20 11:27
    @bigdoggproblem said
    Reasoning:
    I. L<>0 and T<>0 [both are leading digits]
    II. L = 1, because it is a carried digit, and the max (and only) carried digit possible is 1 [9 + 9 = 18].
    III. T + E = LS [column 1; (T + E) must be > 9, since T + a different digit also sums to S, and the only way it can do that is with a carried L]
    IV. H + 1 = T [column 2, given that it receives a carried L] ...[text shortened]... ady = 1, so this is not distinct!

    Therefore, in ALL solutions:
    T = 8; A = 0; H = 7; L = 1
    Biggdoggproblem for the solve!
  7. R
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    30 Jun '20 13:081 edit
    @venda said
    Do we get to know how many possible solutions there are?.
    Bigdogg's logic is sound but I worked it out less logically.
    L has got to be 1 because the total cannot be greater than 1,000
    S therefor cannot be 1 but could be 2,so I tried to make the sum fit by trying the lowest possible total i.e 1022 and so on
    3 solutions in total. I think you just needed to check the possibilities of digits E & S.

    This is how I solved it. Perhaps not very elegant, but I think its thorough.

    1) T + E = 10*x+S.

    Since all digits are 0 -9, "x" may take on two values. namely 0,1 such that the sum ( T+E ) is not greater than 18.

    Start by checking the proposition x = 0

    2) T + E = 10*0 + S = S

    3) E + H = 10*z + S

    by 2):

    E = S - T

    Substitute into 3):

    T - H = 10*z

    Again the placeholder "z" could be 0 or 1. If we check z = 0 then:

    T = H , which is a contradiction as T and H must be distinct.

    Check "z = 1"

    H - T = 10 , again contradiction, max difference between digits is 9.

    So x = 0 is not possible, thus "x" must equal 1 which leads to.

    4) T+E = 10*x+S = 10*1 + S = 10+ S

    5) E + H + x = E + H + 1 = 10*w + S

    Now we can check the possibilities for "w" in a similar manner. Solve 4) for "E". Sub into 5).

    10*(w-1) = H - T+1

    w= 0, contradiction the least the RHS could be is -8

    w = 1 , possible solution:

    6) T = H+1

    So now to the hundreds place ( w= 1):

    L + T + w = L + T + 1 = 10*L + A

    Simplify:

    7) T + 1 = 9*L + A

    We know L <> 0 because its a leading digit, thus L = 1

    8) T + 1 = 9 + A

    9) T = 8 + A

    By substitution of 8) into 9):

    10) H = 7 + A

    From 9) we see "A" has two possible values. A = 0, 1 because 8 ≤ T ≤ 9 .

    However, A = 1 contradicts L = 1 in eq. 7). So we are left with A = 0

    We now have ( using what is above)

    A = 0
    T = 8
    H =7
    L =1

    The two remaining values to find are "E" and "S"

    We know from 5) that:

    E + H + 1 = 10 + S

    Thus ( H =7)

    E + 8 = 10 + S

    E = S + 2

    From here all the possible solutions come from case checking values for E and S against the distinct digit constraint.

    {S,E} = { (0,2) , (1,3) , (2,4) , (3,5) , (4,6) , (5,7) , (6,8) , (7,9) }

    A = 0, ( 1,2 ) not valid
    L = 1, ( 1,3 ) not valid
    H = 7, ( 5,7 ) & ( 7,9) not valid
    T = 8 , ( 6,8 ) not valid

    All solutions ( 3 in total ) are thus:

    A = 0
    L = 1
    H = 7
    T = 8
    {S,E} = { (2,4) , (3,5) , (4,6) }
  8. R
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    30 Jun '20 13:211 edit
    @venda said
    Do we get to know how many possible solutions there are?.
    Bigdogg's logic is sound but I worked it out less logically.
    L has got to be 1 because the total cannot be greater than 1,000
    S therefor cannot be 1 but could be 2,so I tried to make the sum fit by trying the lowest possible total i.e 1022 and so on
    "L has got to be 1 because the total cannot be greater than 1,000"

    Slight correction - I think you mean the total cannot be greater than 2000

    Also, how did you know A = 0 with this method outlined?
  9. R
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    30 Jun '20 13:512 edits
    Also, this problem is not original. It was taken from a problem solving website Brilliant.org. You can go there to find a wide variety of problem solving challenges if this kind of stuff is your thing.

    https://brilliant.org/daily-problems/cryptogram-lass/
  10. Subscribervenda
    Dave
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    30 Jun '20 18:11
    @joe-shmo said
    "L has got to be 1 because the total cannot be greater than 1,000"

    Slight correction - I think you mean the total cannot be greater than 2000

    Also, how did you know A = 0 with this method outlined?
    Yes , I meant 2000.
    I didn't know A was zero, but the lowest total with L equalling 1(as in 1000) and S equalling 2(as in 22) is 1022 which was the first "total" I tried.
  11. R
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    30 Jun '20 18:42
    @venda said
    Yes , I meant 2000.
    I didn't know A was zero, but the lowest total with L equalling 1(as in 1000) and S equalling 2(as in 22) is 1022 which was the first "total" I tried.
    Well at any rate, you found 1 solution. If you would have kept going you would have got all three right in a row. Good work!
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