一條好難的數學題(20pt)

w + x + y + z = 7.11
wxyz = 7.11

w, x, y, z are rational numbers
let w = a/b, x = c/d, y = e/f, z = g/h
consider prime factorization of 711 and 100

aceg/bdfh = 711/100 = 3^2 * 79 / 2^2 * 5^2

note that 79/1, 79/2, 79/4, 79/5, 79/10, 3*79/20, 3*79/25 > 7.11
so one of the item's price can be
79/20 = 3.95
79/25 = 3.16
79/50, 79/100, 3*79/50, 3*79/100

I tried 3.95 for a long time, and failed.

For z = 3.16,
w + x + y = 3.95, wxy = 2.25

Since the factor 79 is taken
a, c, e can take 1, 3, 3 or 1, 1, 9
consider the 1st case (9 looks too large)

Since the factor 25 is taken
b, d, f can take 1, 2, 2 or 1, 1, 4
consider again the 1st case (so many 1 as divisors, looks cannot give us 3.95)

Guess y = 3/2 = 1.5,
w + x = 2.45, wx = 0.375

solve this system of equations
w = 1.2, x = 1.25

Done!
i.e. the prices are 1.2, 1.25, 1.5 and 3.16

2007-03-26 02:18:00 補充：
本來係4個未知數, 2條方程, 應該係有好多答案
但係冇理由seven會賣樣標價係3.141592653589 o既o野...
所以應該係有理數(rational number)
而每一個整數可以寫成一連串質數(prime number)o既積(product)
呢兩點就係解題o既要點
又由於4個未知數+2條方程 o既關係, 難免要估o下
其實可能可以找到o的理由排除某些數, 唔駛亂估亂試

w + x + y + z = 7.11
wxyz = 7.11

w, x, y, z are rational numbers
let w = a/b, x = c/d, y = e/f, z = g/h
consider prime factorization of 711 and 100

aceg/bdfh = 711/100 = 3^2 * 79 / 2^2 * 5^2

note that 79/1, 79/2, 79/4, 79/5, 79/10, 3*79/20, 3*79/25 > 7.11
so one of the item's price can be
79/20 = 3.95
79/25 = 3.16
79/50, 79/100, 3*79/50, 3*79/100

I tried 3.95 for a long time, and failed.

For z = 3.16,
w + x + y = 3.95, wxy = 2.25

Since the factor 79 is taken
a, c, e can take 1, 3, 3 or 1, 1, 9
consider the 1st case (9 looks too large)

Since the factor 25 is taken
b, d, f can take 1, 2, 2 or 1, 1, 4
consider again the 1st case (so many 1 as divisors, looks cannot give us 3.95)

Guess y = 3/2 = 1.5,
w + x = 2.45, wx = 0.375

solve this system of equations
w = 1.2, x = 1.25

Done!
i.e. the prices are 1.2, 1.25, 1.5 and 3.16