Question 1076067: five different positive integers added two at a time give the following sums:16,20,22,23,25,28,29,30,34 and 37.Find the product of the five integers.
Answer by ikleyn(52817) (Show Source):
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1. Let the numbers be a < b < c < d < e (in this order/ordering).
Four pair-wise sums of "a" with four other numbers are among the given set {16,20,22,23,25,28,29,30,34,37}.
Four pair-wise sums of "b" with four other numbers are among the given set {16,20,22,23,25,28,29,30,34,37}.
Four pair-wise sums of "c" with four other numbers are among the given set {16,20,22,23,25,28,29,30,34,37}.
Four pair-wise sums of "d" with four other numbers are among the given set {16,20,22,23,25,28,29,30,34,37}.
Four pair-wise sums of "e" with four other numbers are among the given set {16,20,22,23,25,28,29,30,34,37}.
Therefore, if we sum up all the given numbers, we will get 4*(a+b+c+d+e):
4*(a + b + c + d + e) = 16+20+22+23+25+28+29+30+34+37 = 264.
It implies that
a + b + c + d + e = = 66. (1)
2. Further, 16 is the sum of the two smallest numbers: 16 = a + b. (2)
37 is the sum of the two largest numbers: 37 = d + e. (3)
It implies (by subtracting (2) and (3) from (1)) that c = 66 - 16 - 37 = 13.
So, we just found one of the five numbers, namely, the middle number c = 13.
3. OK. Let's continue our analysis.
So, the two smallest numbers a and b give the sum of 16, and the third number is 13.
It means that the next sum, 20 (see the condition) is a+c.
It can not be nothing else. (Because b + c must be just greater than a + c).
If a + c = 20 and c = 13, then a = 20 - 13 = 7.
It implies b = 16-7 = 9.
So, the first three numbers are a = 7, b = 9 and c = 13.
4. OK. Now we can make the similar analysis from the other end.
So, the two largest numbers d and e give the sum of 37, and the third number in the descending order is 13.
It means that the next from the largest sum, 34 (see the condition) is e+c.
It can not be nothing else. (Because d + c is just less than e + c).
If e + c = 34 and c = 13, then e = 34 - 13 = 21.
It implies d = 37-21 = 16.
5. Thus the numbers are 7, 9, 13, 16 and 21.
Then their product is 275184.
Answer. The product of the numbers is 275184.
Solved.
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