How is it done?

My explanation, of MK's explanation, of EB's brilliant solution of the general 'choose any n from ( n !+ n -1) '.

Like MK I will examine the case n =5 (every essential point may be seen in this case; also it's not too big to handle, and not so small that one can't see the point because so little is going on). Let C 's chosen numbers - between 0 and 123 - be . A will then pass 4 of those 5 cards to P (in a certain order), allowing P to declare the hidden 5th number.

How is it done? First A calculates 2 important numbers ' i ' and ' s ', both of which P may later calculate after the 4 cards are passed:

• i := (mod 5); A holds onto , and passes the other 4 cards
• s := the sum of the other 4 cards (mod 5)

Thus the hidden card satisfies (the vital congruence; 'vital' for reasons that will only be apparent later)

(mod 5) ... (1)

(If you're not fluent with congruences, do a few examples to fix that last important detail: suppose the c 's were 17, 38, 62, 116 and 120. Then

• i := 17+38+62+116+120 = 2+3+2+1+0 = 3 (mod 5); the 'hidden' card is then (namely 116), and
• s is found from s := = 2+3+2+0 = 2 (mod 5)
  

Now, imagine removing the 4 cards from the original 124 numbers (0, 1, 2, 3, ... , 123), and renumber the remaining 120 (=5!) by shifting to the left. In our example

• 0 shift : 0 would still be '0', 1 would still be '1', ... , 16 would still be ' 16 ', while
• 1 shift : 18 would become ' 17 ', 19 would become ' 18 ', ... , 37 would become ' 36 ', and
• 2 shift : 39 would become ' 37 ', 40 would become ' 38 ', ... , 61 would become ' 59 ', and
• 3 shift : 63 would become ' 60 ', 64 would become ' 61 ', ... , 116 (= ) would become ' 113 ', and finally
• 4 shift : 117 would become ' 114 ', ... , 123 would become ' 119 '
  

Thus 116 (= , the hidden card) gets shifted 3 (= i ) places to the left, and thus 's new location is congruent to (- s ) (mod 5). In the general case where ' i ' and ' s ' are defined by

• i := + ... + (mod n ), and the hidden card is
• s := the sum of the other ( n -1) cards (mod n ), and so (mod n )

after renumbering 's new number will be ( ), giving (mod n ). If i =0 then does not get shifted to the left, and its new location - after removing the other ( n -1) c 's - is the same as its earlier one; otherwise gets shifted 1 or 2 or ... or ( n -1) places to the left.

Question . Which numbers (between 0 and 119) are congruent to (- s ) (mod 5)?

Answer, and the point of it all :

• if s =0 they are 5. 0 , 5. 1 , 5. 2 , ... , 5. 22 and 5. 23 , and
• if they are 5. 0 + (5- s ), 5. 1 + (5- s ), 5. 2 + (5- s ), ... , 5. 22 + (5- s ) and 5. 23 + (5- s )

In both cases (and P , knowing the value of s upon receipt of the 4 cards, will know which case is happening) there are exactly 24 (=4!) unknown (to P ) numbers, and A has exactly 24 (=4!) ways in which to pass the 4 cards. THAT'S IT!! All that A and P have to agree upon in advance is a precise correlation between the 4! numbers 0 , 1 , 2 , ... , 23 and the 4! different signals that can be passed with 4 cards, and finally make a use of the congruence (1) above, in order to determine the value of ' i '. The detail is almost an anti-climax: