Le Monde puzzle [#952]

A quite simple Le Monde mathematical puzzle again with Alice and Bob:

In a multiple choice questionnaire with 50 questions, Alice gets a score s such that Bob can guess how many correct (+5 points), incorrect (-1 point) and missing (0 point) Alice got when adding that Alice could not have gotten s-2 or s+2. What is Alice’s score?

A first point is that the overall score is s=5c-i with c+i≤50.  Without further information, the possible results are all integers 0≤c≤50 such that c≥s/5 and 0≤i=s-5c≤50. Possible scores range from -50 to 250, but a quick R check shows that ten values are impossible

for (c in 0:50){
for (i in 0:(50-c))vales[5*c-i+50+1]=1}

which produces

> (1:length(vales))[vales==0]-50-1
[1] 231 236 237 241 242 243 246 247 248 249

Thus looking at the differences, there is only one case for which s-2 and s+2 are impossible values, namely s=239. This means c=48, i=1 since c=49 leads to an impossible i.


6 Responses to “Le Monde puzzle [#952]”

  1. xi’an, your output does show that 245 is an answer, since you’ve got 243 and 247 in your list. You just didn’t “filter” your output completely :-) . 250 is one of those infamous boundary values. (There’s lots of ways to score zero but only one way to score 250)

  2. possible_points <- function(n)
    data.frame(Off = 50-n, Correct = 0:n, Incorrect = n:0, Points = (0:n)*(5) + (n:0)*(-1))

    all_combinations <- dplyr::rbind_all(lapply(1:50, possible_points))
    points <- all_combinations$Points

    dplyr::filter(all_combinations, !(Points + 2) %in% points, !(Points – 2) %in% points)

    # This gives 239, 244, 245, 250

  3. Hi,
    how about 245?

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