+59 Five diffrent postive 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
+6250 \(\text{Let the numbers be }a,b,c,d,e \text{ listed in increasing order}\\ \text{You can get 4 equations straight off}\\ a+b = 16\\ a+c = 20\\ c+e=30\\ d+e=37\)
\(\text{solving this system gets you }\\ (a,b,c,d,e) = (a, 16 - a, 20 - a, 23 - a, 14 + a)\\ \text{You can then create the sums of pairs from this}\\ a+b = 16\\ a+c=20\\ a+d=23\\ a+e = 14+2a\\ b+c=36-2a\\ \text{etc.}\\ \text{plug in the handful of values for }a \text{ that are possible, }1-15\\ \text{sort the results and compare it with the given values}\\ \text{It turns out }a=7 \text{ is the winner, resulting in }\\ (a,b,c,d,e) = (7, 9, 13, 16, 21)\)
\(\left( \begin{array}{cc} \{a,b\} & 16 \\ \{a,c\} & 20 \\ \{b,c\} & 22 \\ \{a,d\} & 23 \\ \{b,d\} & 25 \\ \{a,e\} & 28 \\ \{c,d\} & 29 \\ \{b,e\} & 30 \\ \{c,e\} & 34 \\ \{d,e\} & 37 \\ \end{array} \right)\)
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+129907 Let the integers be (in ascending order) a, b, c, d , e
Then we have these two possible systems
a + b = 16 a + b = 16
a + c = 20 a + c = 20
d + e = 37 d + e = 37
c + e = 34 c + e = 34
a + d = 22 b + c = 22
In the first system we have that
e - a = 14 and
e - a = 15
impossible
In the second system we have that
c = 22 - b
And
a + b = 16 (1)
a + (22 - b) = 20 ⇒ a - b = - 2 (2)
Adding (1) and (2) we have that 2a = 14 ⇒ a = 7
So
b = 9
c = 22 - 9 = 13
e = 34 - c = 34 - 13 = 21
d = 37 - e = 37 - 21 = 16
So....the five integers are
7, 9, 13, 16, 21
