Pour les matheux:Richard Green

The four fours problem is to find a mathematical expression for a specific number N that involves four occurrences of the digit 4, and no other digits. The picture shows a possible way to express each of the numbers 0 up to 20 using four 4s. Recall that “4!” (4 factorial) means 4 x 3 x 2 x 1, or 24.

It turns out that the four fours problem can be solved for all the numbers from 0 to 100, but some of these require some ingenuity. The hardest is probably 73, which is equal to (4! C √(4))/4 + 4, where (n C r) = n!/(r! (n-r)!) is the binomial coefficient operation. In this case, 4! C √(4) = 24 C 2 is equal to (24!/(2! x 22!)), which cancels down to (24 x 23)/2, or 276. Dividing this by 4 and then adding 4 yields 73, as desired.

The use of logarithmic operators is usually banned in this problem, since there is a general method to produce any positive integer N using four 4s, four logarithms, N+1 square root signs, a minus sign, two divisions and two sets of parentheses. The full logarithmic formula appears on Wikipedia (http://en.wikipedia.org/wiki/Four_fours) and a solution of the problem from 0 up to 100 may be found here (http://mathforum.org/ruth/four4s.puzzle.html).

Thanks to +Carl Niclas for helping me with this post.

#mathematics