Logical Solutions

Great problem, @Anish Harsha -- Thanks for sharing it!

To everyone -- there are a few methods to solve this. How did you solve it? Post your solution!

Here's one method: Note that the three digit numbers are in the form $100A + 10A + A = 111A,$ so we have $111(A+B+C) = 111\cdot D \quad \iff \quad A+B+C = D.$

Let's fix a value for $D.$ Then, for any combination of unique digits that sum to $D$ there are $3!=6$ possibilities for $(A,B,C).$

If $D =6,$ the only possibility is $1+2+3.$

If $D=7,$ the only possibility is $1+2+4,$ since $1+3+3$ would not give unique digits.

If $D=8,$ we can have $1+2+5$ or $1+3+4$ .

If $D=9,$ we can have $1+2+6$ or $1+3+5$ or $2+3+4$ .

Thus, there are a total of $6(1+1+2+3) = 6\cdot 7 = \boxed{42}$ unique solutions to this cryptogram.

Since each line is divisible by 111, this is equivalent to finding the number of different solutions to A+B+C=D, where A, B, C, D are distinct single digit values. There are 7 possible choices for the set {A, B, C}, whose sum does not exceed 9, namely {1,2,3}, {1,2,4}, {1,2,5}, {1,2,6}, {1,3,4}, {1,3,5}, {2,3,4}. Each of these has 6 possible orderings for the three digits, giving 42 solutions. How does this qualify as level 4? It seems a little straightforward...

No carry forwards and only positive integers:

1
2
3
4
5
6

1   1   1   1   1   1   2   
2   2   2   2   3   3   3   
3   4   5   6   4   5   4   
6   7   8   9   8   9   9   

6   6   6   6   6   6   6   42

3! permutations for each. Total = 6 $\times$ 7 = 42

Answer: $\boxed{42}$

Nice problem. First when all are unique digits, and the end sum cannot have a carry. So the numbers that sum to a single digit when 3 are selected from them in any combo are 1,2,3,4. Then the other 3 digit combos that give a single digit summation are 512, 513 and 612. When considering other higher numbers as part of the summation, the 3numbers cannpt be unique, positive and result in single digit sum. So for 1,2,3,4 number of 3 set selections can be made are 4c3 -> 4. For each 3set, permutations is 6. So (4+3)*6 -> 42

It has to be 9 $\ge$ A + B + C and A, B, C have to be distinct $\Rightarrow$ the posibilities are (A,B,C) = (1,2,3), (1,2,4), (1,2,5) , (1,2,6), (1,3,4),(1,3,5) or (2,3,4) and there are 6 permutations for each one of these posibilities. Total = 42 ways