Решение задачи о двух мудрецах и числах от 1 до 100

import Data.List

groupedSet :: (Num a, Eq a, Ord a) => (a -> a -> a) -> [a] -> [[(a, a)]]
groupedSet op xs =
    map (map snd) . groupBy (\(a, _) (b, _) -> a == b) . sort $
        [(x `op` y, (x, y)) | x <- xs, y <- xs, x <= y]

main :: IO ()
main = do
    let a = groupedSet (+) numbers
        {- i do not know these numbers: -}
        b = filter ((> 1) . length . take 2) $ groupedSet (*) numbers
        {- i know that you do not know them: -}
        c = [x | x <- a, all (`elem` concat b) x]
        {- then i know these numbers: -}
        d = singlets b $ concat c
        {- then i know them too: -}
        e = singlets a $ concat d
    print $ concat e
    where numbers      = [2 .. 99]
          singlets a b = [y | x <- a, let y = b `intersect` x, length y == 1]

Для решения я использовал MATLAB. Да, я стреляю из пушки по воробьям, но с Матлабом я часто сталкиваюсь в работе

numbers = Union[Sort /@ Tuples[Range[2, 99], 2]];
goodproducts =   Select[Reap[Sow[Plus @@ ##, Times @@ ##] & /@ numbers, _, List][[2]], (Length[##[[2]]] >= 2) &];
goodsums =   Select[Reap[Sow[Times @@ ##, Plus @@ ##] & /@ numbers, _,  List][[2]], (Complement[##[[2]], First /@ goodproducts] == {}) &];
goodproducts =   Select[goodproducts, (Length[Intersection[##[[2]], First /@ goodsums]] == 1) &];
goodsums =  Select[goodsums, (Length[Intersection[##[[2]], First /@ goodproducts]] == 1) &];
Select[numbers, And[MemberQ[First /@ goodproducts, Times @@ ##],  Plus @@ ## == goodsums[[1, 1]]] &][[1]]

Поначалу я не знал, когда у Шерил день рождения, но знаю теперь.

using System;
using System.Linq;
class Program
{
	static void Main()
	{
		var birthDays = Enumerable.Range(2, 98)
			.Join(
				Enumerable.Range(2, 98), n => 1, n => 1,
				(n1, n2) => new {D = n1*n2, M = n1 + n2})
			.Distinct().ToList();
		var x1 = birthDays.Where(d => !birthDays.Any(d1 => d.D == d1.D && d.M != d1.M)).ToList();
		var x2 = birthDays.Where(d => x1.All(d1 => d1.M != d.M)).ToList();
		var x3 = x2.Where(d => !x2.Any(d1 => d1.D == d.D && d1.M != d.M)).ToList();
		var solve = x3.Where(d => !x3.Any(d1 => d1.D != d.D && d1.M == d.M)).ToList();
		Console.WriteLine(string.Join(", ", solve.Select(d => d.D + " " + d.M)));
	}
}