16 ARITHMETICAL RECREATIONS [CH. I 



which he reckons as (n + p)th is on the hour (20 — p). Hence, 

 putting p = 20 — n, the tap which he reckons as 20th is on the 

 hour n. Of course the hours indicated by the first seven taps 

 are immaterial : obviously also we can modify the presentation 

 by beginning on the hour viii and making 21 consecutive taps, 

 or on the hour IX and making 22 consecutive taps, and so on. 



Third Example. The following is another simple example. 

 Suppose that a pack of n cards is given to some one who is asked 

 to select one out of the first m cards and to remember (but not to 

 mention) what is its number from the top of the pack ; suppose 

 it is actually the #th card in the pack. Then take the pack, 

 reverse the order of the top m cards (which can be easily 

 effected by shuffling), and transfer y cards, where y<n — m, 

 from the bottom to the top of the pack. The effect of this is 

 that the card originally chosen is now the (y + m — a; + l)th 

 from the top. Return to the spectator the pack so rearranged, 

 and ask that the top card be counted as the (x + l)th, the next 

 as the (x + 2)th, and so on, in which case the card originally 

 chosen will be the (y + m + l)th. Now y and m can be chosen 

 as we please, and may be varied every time the trick is per- 

 formed ; thus any one unskilled in arithmetic will not readily 

 detect the method used. 



Fourth Example*. Place a card on the table, and on it 

 place as many other cards from the pack as with the number 

 of pips on the card will make a total of twelve. For example, 

 if the card placed first on the table is the five of clubs, then 

 seven additional cards must be placed on it. The court cards 

 may have any values assigned to them, but usually they are 

 reckoned as tens. This is done again with another card, and 

 thus another pile is formed. The operation may be repeated 

 either only three or four times or as often as the pack will 

 permit of such piles being formed. If finally there are p such 

 piles, and if the number of cards left over is r, then the sum 

 of the number of pips on the bottom cards of all the piles will 

 be 13(p-4) + r. 



For, if a; is the number of pips on the bottom card of a pile, 

 * A particular case of this problem was given by Bacaet, problem xvn, p. 138. 



