[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.**

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

ReplyDeleteHow? Please show us.

Refer to http://www.hci.sg/aphelion/apmops/2007/pdf/Chinese/2006%20Chinese%20IR.pdf

ReplyDeleteIt's adapted from Q3 from SAPMOPS 2006 Invitational round.

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

ReplyDelete