兩條數學難題

1) This is the famous 7-11 Problem. The prices are: $1.20, $1.25, $1.50, and $3.16
We have such equation:
x y z w = 7.11
x + y + z + w = 7.11

If x,y,z,w represent the amounts of the four items in cents then we have
x y z w = 711000000 cents
x + y + z + w = 711 cents

Note that the factors : 711 = (1)(3)(3)(79)

If there is to be an exact solution then the product xyzw must be divisible by 3, 9, 79.
The exact sum of the 4 items is 7.11 so the cost of each item is less than $7.11.

So try and error will show $3.16 (= 79*0.04), $1.50, $1.25, $1.20 is the required one.

2) IT IS DEFINITELY NOT 1/2!!
This is one common misconception that 幾何級數幾時都o岩.... Remember, geometric series only true when the factor is |a|<1
so in this case a = -1 does not work.

Usually people make this paradox by saying:
S = 1-1+1-1+...
S-1 = -1+1-1+... = -S
so 2S-1=0
=> S = 1/2

Why this is false? Because in the first place S is not defined. 定義唔到o既數係唔可以囉黎計.

Let me give you an example: 1-1+1-1+1....
we can interchange the 2/3 terms, 4/5 terms, 6/7 terms and so on and write it as :
1+(1-1)+(1-1)+(1-1)+...
= 1+0+0+0+0+...
=1

You see the problem? Now the sum is 1, it cannot be 1/2....
anyway, a not well defined sum does not exist.

Mathematicians call the sequence 1-1+1-1+1.... "DIVERGENT " 發散無窮級數

2006-12-13 09:51:09 補充：
幾何級數幾時都o岩I mean the formula Sum a^k = 1/1-a

只回答第二題.

如果把這個視為幾何系數 (倍數小於 1) 的和 (如: 1 + 1/2 + 1/4 + 1+8 + ... 下個數字是前面數字的 1/2), 那麼根据方程式.

sum = 1 / (1 - a), a 是倍數 (上例是 1/2)

你的題目中, a = (-1).

但, 其實上面的方程式只應該應用於倍數小於 1 的情況.