kalisditz (2001-03-29) [problem from Prof. Doug Brumbaugh, UCF]

「求四件東西各自的價錢,這四件東西相乘等於7.11,相加也等於7.11。」

原問題:In a "seven-eleven" (7-11) store, a customer selected four items to buy. The check-out clerk says that he multiplied the costs of the items and obtained exactly 7.11, the very name of the store! The customer calmly tells the clerk that the costs of the items should be added, not multiplied. The clerk then informs the customer that the correct total is also $7.11.
What are the exact costs of the 4 items?

答案:
7.11 = 3.16 + 1.25 + 1.50 + 1.20 = 3.16 x 1.25 x 1.50 x 1.20
那四件東西的價錢分別就是是:3.16 + 1.25 + 1.50 + 1.20 (唯一的解答)

 

解釋(暫時只翻譯了一部分):

假設a,b,c及d是那個四件物品的價錢,為了方便運算和解釋,在此使用cent(仙)為單位,而不是dollar(元)。
於是得出以下算式:

1) a+b+c+d = 711
2) abcd = 711000000 = 26 x32x56 x79.

由此可推斷,其中一件東西的價錢必定是79的倍數。
價格只可能是兩位小數,由於79乘以9已等於711,所以這個倍數一定只能少於9,即是1,2,3,4,5,6或8,共七個可能性(7x79等於553,711000000無法被553整除,所以這個可能性被否決)。

倘若用公式來表達,這七個可能性就是:

1. a=79, b+c+d=632, bcd=9000000
2. a=158, b+c+d=553, bcd=4500000
3. a=237, b+c+d=474, bcd=3000000
4. a=316, b+c+d=395, bcd=2250000
5. a=395, b+c+d=316, bcd=1800000
6. a=474, b+c+d=237, bcd=1500000
7. a=632, b+c+d=79, bcd=1125000

Now, the product of 3 positive numbers of given sum is greatest when they are all equal, which means that the product bcd cannot exceed (b+c+d)3/27. This rules out the last three of the above 7 cases.

In the first three cases, on the other hand, the sum b+c+d isn't a multiple of 5, so at least one of b,c,d (say d) isn't either. Therefore, the product bc must be a multiple of the sixth power of 5. Since neither b nor c can be large enough to be a multiple of the fourth power [625 is clearly too big a share of 711, leaving only 7 cents for two items in the first case and nothing at all in the other two] we must conclude that both b and c are nonzrero multiples of 125. The number d would thus be obtained by subtracting from one of three possible sums (632, 553, 474) some multiple of 125 (necessarily: 250, 375 or [in the first two cases] 500). This gives only 8 possible values for d (382, 257, 132, 303, 178, 53, 224, 99). Each is unacceptable because it has a prime factor other than 2 or 3.

Therefore, only the fourth case is not ruled out, so that a=$3.16.
a=316, b+c+d=395, bcd=2250000.

Since the sum b+c+d is a multiple of 5, and at least one of the terms is a multiple of 5, either only one is, or all are. The former option is ruled out since this would imply for the single multiple of 5 to be a multiple of 56=15625, which would, by itself, be much larger than the entire sum of 395. So, b, c, and d are all multiples of 5, and we may let b=5b', c=5c', d=5d', where b'+c'+d'=79 and b'c'd'=18000.

Now, these three new variables may not be all divisible by 5 (otherwise their sum would be too). It's not possible either to have a single one of them divisible by 5, because it would then have to be a multiple of 53=125 and exceed the whole sum of 79. Therefore, we must have a multiple of 25 (say b'=25b") and a multiple of 5 (say c'=5c"): 25b"+5c"+d'=79 and b"c"d'=144. It is then clear that b" can only be 1 or 2. If b" was 2, then we would have 5c"+d'=29 and c"d'=72, implying that c" is a solution of x(29-5x)=72 or 5x2-29x+72=0. However, this quadratic equation does not have any real solutions, because its discriminant is negative. Therefore, b" is equal to 1 and b=5(25b")=125=$1.25.

The whole thing thus boils down to solving 5c"+d'=54 and c"d'=144, which means that c" is solution of x(54-5x)=144 or 5x2-54x+144=0. Of the two solutions of this quadratic equation (x=6 and x=4.8), we may only keep the one which is an integer. Therefore c"=6, c'=30, c=150=$1.50.

Finally, d'=144/c=24 which means d=5d'=120=$1.20.

Therefore, the solution is unique (the order of the 4 items being irrelevant):

$3.16, $1.25, $1.50, $1.20.