Tuesday, June 22, 2010

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.
-------------------------------------------------
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.

4 comments:

  1. Is your method to try from 5 to 3120? But obviously you cannot do that in exam conditions. It is possible to solve it without calculator and not time consuming.

    ReplyDelete
  2. 'It is possible to solve it without calculator and not time consuming.'

    How? Please show us.

    ReplyDelete
  3. Refer to http://www.hci.sg/aphelion/apmops/2007/pdf/Chinese/2006%20Chinese%20IR.pdf
    It's adapted from Q3 from SAPMOPS 2006 Invitational round.

    ReplyDelete
  4. XIAO GAO they dare come up with such a killer qs?

    ReplyDelete