CARD TRICK OR MATH TREAT
The card trick
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21 cards are taken from which viewer selects one card
The 21 cards are divided into 3 piles
Viewer identifies pile in which selected card is present
That pile is kept below all other piles and 21 cards are
divided into 3 piles again(left to right distribution)
• Process is repeated 3 more times
• Second card from top is taken in last step
• IT IS THE SELECTED CARD!!
DEFINITION OF VARIABLES
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N  Total no. of cards
C  no. of cards in each pile
w Position of the card in the pile (from top)
P  total no. of piles
S  no. of steps
1) First position of card ‘A’ is ‘w’ in pile ‘k’
2) one division - pile ‘k’ is kept below all other
piles and N cards are divided into ‘P’ piles
What is the position of card ‘A’ now in the pile?
It is [ (C –w) / P] + 1
• if Card ‘A’ is in ‘w’ position from top in pile ‘k’
then there are (C-w) cards below it .
new pos
P
…
3 2 1 (1)
2P
…
P+1 (2)
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(m+1)P
.. (C-w) ….
mP+1 (m+1)
So if C-w lies in (m+1) row , not extreme card,
new posm +1 [(C-w)/P] +1
If C-w = (m+1)P then new posm+2[(C-w)/P] +1
FUNCTION DEFINITION
f : {1,2,…C}{1,2….C} such that
f(w) = [( C– w)/P]+1
PROPERTIES OF FUNCTION
1) f is decreasing function
2) f(a) = a and f(b)=b  a = b
3) if x(1)x(2)x3…..x(n)x(1)
(all x(i) distinct) then n = 1 or 2 only
Case1) x(1)<x(2) and x(1)<x(3)
Position (inc order)
(position after div)
x(1) ..
x(2) ..
x(3)..
x(4) ..
x(n)..
x(1) ..
x(4)..
x(5) ..
x(2)..
x(3) ..
Col 1-inc Col 2 -dec ( f is dec)
so x(3)<x(1). Contradiction . Similarly other cases proved
4)f(x)=y and f(y)=x (x!=y) f(a)!=a for any ‘a’
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proof: suppose there is ‘a’ where f(a)=a.
wlog we can assume x<y
since f is decreasing,x<a<y is the only possibility.
[(C-x)/p]+1=y & [(C-y)/p]+1=x & [(C-a)/p]+1=a
C=(y-1)p+x+k1=(x-1)p+y+k2=(a-1)p+a+k3 (for some ki in
Z)
(y-x)p=y-x+k1-k2 =>(y-x)(p-1)=k1-k2
LHS>0 =>p-1/k1-k2 this is possible if k1=p-1 and k2=0
y=x+1
But x<a<y .so contradiction arises.
5) the no. of steps are infinite iff f(a)!=a for any ‘a’
1) ( if f(a) != a then infinite no. of steps)
1 x  y  z….
C 1  x  y….. So always, at any step f(1)!=f(C)
2) ( if infinite no. of steps then f(a) != a)
if there exist x, y x-->y-->x then from (4) f(a)!=a
if no such x,y exists then from 3, established that
no cycle > 2 is present. Hence system is finite
6) if f(a)=a ,the final position occupied by the card is ‘a’
proof: f(a)=a  xyx doesn’t exist from (4,5)
if any system x(0)x(1)…..x(C-1) .. Is finite then
f (x(C-1)) = x(C-1) = a ( from 2)
hence all no.s 1,2,3,4…C finally go to ‘a’
thus ‘a ‘is the final position
RESULTS
1)The card trick can be performed if f(a)=a for
some ‘a’ belonging to {1,2,3..C}
2)final position of card will be ‘a’
3)Comp prog given below prints no. of steps
and the final position of the card
SOME EXAMPLES
P
3
7
6
3
C
5
8
2
4
final pos
2
1
1
steps
3
2
infi
2