SAPMOPS 2006

Abel, Bob, Conan, Dave and Elijah divided a certain number of marbles amongst themselves in the following way: Abel took 1 marble and 20% of the remaining marbles, then Bob took 1 marble and 20% of the remaining marbles. Conan, Dave and Elijah did the same. At least how many marbles were there in tbe beginning?


[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + 5/4 (1 + 5/4 (1 + 5/4 (1 + R)))))    units not drawn to scale
[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + 5/4 (1 + 5/4 (1 + R))))       units not drawn to scale
[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + 5/4 (1 + R))   units not drawn to scale

[1][ ][  ][  ][  ][  ] --> 1 + 5/4 (1 + R)
        [1][][][][][] --> 1 + R


As R is a multiple of 5, try till you get the first expression as a whole number. Solve using Microsoft Office - Excel, the first whole number is 3121.

There were 3121 marbles in the beginning.


Alternative solution as suggested by a User.
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Let x be the number of marbles in the beginning.

[                                                             x - 1                                                              ][1]
[                                                      4/5 x - 4/5                                                        ][ A  ]
[                                                   4/5 x - 4/5 - 1                                                 ][1]     
The remaining marbles after A has taken his share.


[                                        (4/5)^2 x - (4/5)^2 - 4/5                                       ][ B ]
[                                   (4/5)^2 x - (4/5)^2 - (4/5) - 1                                ][1] 
The remaining marbles after A and B have taken their shares.


[                        (4/5)^3 x - (4/5)^3 - (4/5)^2 - (4/5)                            ][  C  ]
[                     (4/5)^3 x - (4/5)^3 - (4/5)^2 - (4/5) - 1                     ][1]
The remaining marbles after A, B and C have taken their shares.


[              (4/5)^4 x - (4/5)^4 - (4/5)^3 - (4/5)^2 - (4/5)             ][  D  ]
[         (4/5)^4 x - (4/5)^4 - (4/5)^3 - (4/5)^2 - (4/5) - 1        ][1]
The remaining marbles after A, B, C and D have taken their shares.


[(4/5)^5 x - (4/5)^5 - (4/5)^4 - (4/5)^3 - (4/5)^2 - (4/5) ][   E  ]

The remaining marbles, r --> (4/5)^5 x - (4/5)^5 - (4/5)^4 - (4/5)^3 - (4/5)^2 - (4/5)

The formula for goemetric series (which is taught at Sec 3) is as follows:
1 + z + z^2 + z^3 + . . . + z^n = [z^n+1 - 1]/[z - 1]

(4/5) + (4/5)^2 + (4/5)^3 + (4/5)4 + (4/5)^5 = [(4/5)^6 - 1]/[4/5 - 1] - 1

= - 5(4/5)^6 + 5 - 1

= - 4 (5/4) (4/5)^6 + 4

= - 4 (4/5)^5 + 4

r --> (4/5)^5 x + 4 (4/5)^5 - 4 = (x + 4) (4/5)^5 - 4

For r to be a whole number, (x + 4) must be equal or greater than 5^5 

For the smallest r, (x + 4) = 5 x 5 x 5 x 5 x 5
                                       x = 3125 - 4
                                          = 3121 marbles.

There were 3121 marbles in the beginning.

It is obvious that this cannot be P5/6 Stuff.