Just paste that code into [1] and append the line (or place it at the start, doesn't matter)

    :- initialization(colin_score(X)).

[1]: https://quantumprolog.sgml.net/browser-demo/browser-demo.htm...

[2]: https://morepablo.com/2010/09/some-professor-layton-prolog.h...

I would also challenge the author to consider puzzles a bit more. Sprinkling fun into the topics goes a long way. Specifically, do both! Talk about how you can reframe a "real world" problem in terms of some puzzles we may have played as kids. Invite us to play more as adults.

1) there are 4 questions that everyone answered the same way (let's say this earned everyone A points)

2) in the rest of the questions, 7-point and 5-point students gave opposite answers

that means that 7-A (rest of 7-point student's correct answers) equals to 5-(4-A) (rest of 5-point student's incorrect answers)

7-A = 1+A => A=3

That means 3-point student scored 3 points together with everyone else, and answered rest of the questions wrong

From this it's trivial to find the hidden score and why there are 4 possible answer keys

    import z3
    res = [("bbababbabb", 7), ("baaababaaa", 5), ("baaabbbaba", 3), ("bbaaabbaaa", z3.Int("x"))]
    key = [z3.Int(f"k{i}") for i in range(10)]
    z = z3.Solver()
    z.add([z3.Sum([ord(a) == k for a, k in zip(r[0], key)]) == r[1] for r in res])
    z.check()
    print(z.model()[res[-1][1]])

Anecdotally, some years ago the Zebra Puzzle [1] made the rounds in my team. Two people solved it: myself, a young intern, who mapped out the constraints as a physical puzzle that I was able to solve visually, and a more seasoned colleague who used Prolog.

[1] https://en.wikipedia.org/wiki/Zebra_Puzzle

Heuristics could reduce trial-and-error to a reasonable search space, but is there a purely deductive way to solve this by hand?

    :-use_module(library(clpfd)).

    % score/3 has mode ++- and maps
    % a given answer and a correct answer to a score.
    score(Given, Correct, Score) :-
        zcompare(Order, Given, Correct),
        score_(Order, Score).

    score_(=, 10).   % 10 points if given matches correct,
    score_(<, 0).    % else 0 points
    score_(>, 0).

    puzzle(ColinTotal, ColinScores, CorrectAnswers) :-
        A = 0,                      % false
        B = 1,                      % true
        length(CorrectAnswers, 10), % ten unknown correct answers,
        CorrectAnswers ins 0..1,    % all either true or false.

        % compare player answers with correct answers, add their scores.
        maplist(score, [B,B,A,B,A,B,B,A,B,B], CorrectAnswers, Mary),
        sum(Mary, #=, 70),
        maplist(score, [B,A,A,A,B,A,B,A,A,A], CorrectAnswers, Dan),
        sum(Dan, #=, 50),
        maplist(score, [B,A,A,A,B,B,B,A,B,A], CorrectAnswers, Lisa),
        sum(Lisa, #=, 30),
        maplist(score, [B,B,A,A,A,B,B,A,A,A], CorrectAnswers, ColinScores),
        sum(ColinScores, #=, ColinTotal).

    ?- puzzle(ColinTotal, ColinScores, CorrectAnswers).

    ColinTotal = 60,
    ColinScores =    [10, 10, 10,  0, 10,  0, 10,  0, 10,  0],
    CorrectAnswers = [ 1,  1,  0,  1,  0,  0,  1,  1,  0,  1]

    ColinTotal = 60,
    ColinScores =    [10, 10, 10,  0, 10,  0,  0, 10, 10,  0],
    CorrectAnswers = [ 1,  1,  0,  1,  0,  0,  0,  0,  0,  1]

    ColinTotal = 60,
    ColinScores =    [10, 10,  0,  0, 10,  0, 10, 10, 10,  0],
    CorrectAnswers = [ 1,  1,  1,  1,  0,  0,  1,  0,  0,  1]

    ColinTotal = 60,
    ColinScores =    [ 0, 10, 10,  0, 10,  0, 10, 10, 10,  0],
    CorrectAnswers = [ 0,  1,  0,  1,  0,  0,  1,  0,  0,  1]

[0]: https://sentient-lang.org/

  %%% This works in swi as is
  %%% in ciao we need to include some modules
  %:- use_module(library(numlists)).
  %:- use_module(library(hiordlib)).
  %:- use_package(clpfd).

  main :-
   M = [b,b,a,b,a,b,b,a,b,b],
   D = [b,a,a,a,b,a,b,a,a,a],
   L = [b,a,a,a,b,b,b,a,b,a],
   C = [b,b,a,a,a,b,b,a,a,a],
   score(K,M,7), score(K,D,5), score(K,L,3),
   score(K,C,S),
   write(S). 
  
  score(K,T,S) :-
   maplist(eq_OZ,K,T,SL),
   sum_list(SL,S).
  
  eq_OZ(X,X,1).
  eq_OZ(X,Y,0) :- dif(X,Y).
  
  
  %%% ciao doesn't have dif/2 so we use this
  %eq_OZ(X,Y,0) :- X #\= Y.