Maths(HKMO)

1. Consider the following equivalent problem:
There are 12 O's in a line, find the number of ways to insert 3 X's so that each X lies between two consecutive O's.

OOOOXOOXOOOXOOO 12=4+2+3+3
OOXOOOXOOOOOOXO 12=2+3+6+1

As there are 11 slits between the O's, the answer is 11C3=165
W=165

2. The radius of the circle is 4m/(2Pi)=7/11 m
Area of the hexagon=total area of 6 equilateral triangles with side 7/11 m
=6(1/2)(7/11)(7/11)sin(Pi/3)=147sqrt(3)/242 m^2
A=147sqrt(3)/242

3. 7^2006=(7^4)^501*7^2=(2401)^507*49
C=9

1. w, x, y and z must be of values from 1 - 9, since sum of them = 12.

Possible solutions are as follows:-

{1,1,1,9} = 4P4 / 3P3 = 4
{1,1,2,8} = 4P4 / 2P2 = 12
{1,1,3,7} = 4P4 / 2P2 = 12
{1,1,4,6} = 4P4 / 2P2 = 12
{1,1,5,5} = 4P4 / (2P2 x 2P2) = 6
{1,2,2,7} = 4P4 / 2P2 = 12
{1,2,3,6} = 4P4 = 24
{1,2,4,5} = 4P4 = 24
{1,3,3,5} = 4P4 / 2P2 = 12
{1,3,4,4} = 4P4 / 2P2 = 12
{2,2,2,6} = 4P4 / 3P3 = 4
{2,2,3,5} = 4P4 / 2P2 = 12
{2,2,4,4} = 4P4 / (2P2 x 2P2) = 6
{2,3,3,4} = 4P4 / 2P2 = 12
{3,3,3,3} = 1

W = 153

2.
2(Pi)r = 4
2(22/7)r = 4
r = 7/11 m

A = 6 x {1/2 x (7/11)(7/11) sin 60} = 147/242 sqrt(3)


3.

Since 7^4 = 2401

2006 = 501 x 4 + 2

C = 9

2007-01-11 23:34:18 補充：
Sorry, W has been mistakenly calculated above. It should be 165. Thanks to Miao Ming.