CH. XI] MISCELLANEOUS PROBLEMS 243 



Suppose that the pile now indicated as containing the 

 selected card is taken up 6th: then (i) at the top of the 

 pack are 9 (6 — 1) cards ; (ii) next are 9 — 3a cards ; (iii) next 

 are 3 cards, of which one is the selected card ; and (iv) lastly 

 are the remaining cards of the pack. The cards are dealt out 

 now for the third time : in each pile the bottom 3 (6 — 1) cards 

 will be taken from (i), the next 3 — a cards will be taken from 

 (ii), the next card will be one of the three cards in (iii), and the 

 remaining 8 - 36 + a cards are from (iv). 



Hence, after this deal, as soon as the pile is indicated, it 

 is known that the card is the (9 — 36 + a)th from the top 

 of that pile. If the process is continued by taking up this 

 pile as cth, then the selected card will come out in the place 

 9 (c — 1) + (8 — 36 + a) + 1 from the top, that is, will come out 

 as the (9c — 36 + a)th card. 



Since, after the third deal, the position of the card in the 

 pile then indicated is known, it is easy to notice the card, in 

 which case the trick can be finished in some way more effective 

 than dealing again. 



If we put the pile indicated always in the middle of the 

 pack we have a = 2, 6 = 2, c = 2, hence n = 9c — 36 + a = 14, 

 which is the form in which the trick is usually presented, as 

 was explained above on page 241. 



I have shown that if a, 6, c are known, then n is determined. 

 We may modify the rule so as to make the selected card come 

 out in any assigned position, say the nth. In this case we 

 have to find values of a, 6, c which will satisfy the equation 

 n = 9c — 36 + a, where a, 6, c can have only the values 1, 2, 

 or 3. 



Hence, if we divide n by 3 and the remainder is 1 or 2, this 

 remainder will be a; but, if the remainder is 0, we must decrease 

 the quotient by unity so that the remainder is 3, and this 

 remainder will be a. In other words a is the smallest positive 

 number (exclusive of zero) which must be subtracted from n 

 to make the difference a multiple of 3. 



Next let p be this multiple, i.e. p is the next lowest integer 

 to n/3 : then Sp = 9c — 36, therefore p = 3c — 6. Hence 6 is 



16—2 



