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My four numbers (Posted on 2016-02-24) Difficulty: 3 of 5
I have in mind 4 positive numbers, which I would like you to find.

Out of the set of 6 pair-wise products of those numbers I am going to omit one.

The remaining five are: 8,10,12,24,30.

See The Solution Submitted by Ady TZIDON    
Rating: 4.3333 (3 votes)

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Solution Solution with Analysis | Comment 3 of 12 |
Begin by finding the omitted pairwise product.  There are 10 ways to make products from the 5 given values.  The product of all four original numbers must occur twice.

8*10 = 80
8*12 = 96
10*12 = 120
8*24 = 192
8*30 = 240
10*24 = 240
12*24 = 288
10*30 = 300
12*30 = 360
24*30 = 720

240 appears twice and uses 8, 10, 24, and 30.  Therefore the sixth pairwise product is 240/12 = 20.

Let the four original numbers be A,B,C,D in ascending order.  Now that all six pairwise products are known, A*B equals the smallest product: 8, A*C equals the second smallest: 10, C*D equals the largest product: 30, and B*D equals the second largest product 24.  Finally A*D and B*C equals 12 and 20 in some order.   In summary, the numbers solve the system:
A*B = 8
A*C = 10
B*D = 24
C*D = 30
A*D = 12 -OR- A*D = 20

The first four equations resolve to make B=8/A, C=10/A, and D=3*A.  Case 1 with A*D=12 yields solutions (A,B,C,D) = (2, 4, 5, 6) or (-2, -4, -5, -6).  Case 2 with A*D = 20 yields solutions (A,B,C,D) = (2*sqrt(5/3), 4*sqrt(3/5), 5*sqrt(3/5), 6*sqrt(5/3)) or (-2*sqrt(5/3), -4*sqrt(3/5), -5*sqrt(3/5), -6*sqrt(5/3))

Two of these solutions satisfy the positive criterion: (A,B,C,D) = (2, 4, 5, 6) and (2*sqrt(5/3), 4*sqrt(3/5), 5*sqrt(3/5), 6*sqrt(5/3))

If the assumption is made that the values must be positive integers then there is only one solution: (A,B,C,D) = (2, 4, 5, 6)

  Posted by Brian Smith on 2016-02-24 23:59:05
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