Posers and Puzzles

Posers and Puzzles

  1. Subscriberjoe shmo
    Strange Egg
    podunk, PA
    Joined
    10 Dec '06
    Moves
    8218
    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
    S.Yorks.England
    Joined
    18 Apr '10
    Moves
    67607
    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. Subscriberjoe shmo
    Strange Egg
    podunk, PA
    Joined
    10 Dec '06
    Moves
    8218
    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. SubscriberBigDoggProblem
    Not entirely stable
    But close enough
    Joined
    26 Nov '04
    Moves
    134459
    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
    S.Yorks.England
    Joined
    18 Apr '10
    Moves
    67607
    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. Subscriberjoe shmo
    Strange Egg
    podunk, PA
    Joined
    10 Dec '06
    Moves
    8218
    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. Subscriberjoe shmo
    Strange Egg
    podunk, PA
    Joined
    10 Dec '06
    Moves
    8218
    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. Subscriberjoe shmo
    Strange Egg
    podunk, PA
    Joined
    10 Dec '06
    Moves
    8218
    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. Subscriberjoe shmo
    Strange Egg
    podunk, PA
    Joined
    10 Dec '06
    Moves
    8218
    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
    S.Yorks.England
    Joined
    18 Apr '10
    Moves
    67607
    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. Subscriberjoe shmo
    Strange Egg
    podunk, PA
    Joined
    10 Dec '06
    Moves
    8218
    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!
Back to Top