 # Let the Lass... joe shmo Posers and Puzzles 29 Jun '20 17:05
1. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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!