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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

 Feb 25, 2019
 #1
avatar+6045 
+1

\(\text{Let the numbers be }a,b,c,d,e \text{ listed in increasing order}\\ \text{You can get 4 equations straight off}\\ a+b = 16\\ a+c = 20\\ c+e=30\\ d+e=37\)

 

\(\text{solving this system gets you }\\ (a,b,c,d,e) = (a, 16 - a, 20 - a, 23 - a, 14 + a)\\ \text{You can then create the sums of pairs from this}\\ a+b = 16\\ a+c=20\\ a+d=23\\ a+e = 14+2a\\ b+c=36-2a\\ \text{etc.}\\ \text{plug in the handful of values for }a \text{ that are possible, }1-15\\ \text{sort the results and compare it with the given values}\\ \text{It turns out }a=7 \text{ is the winner, resulting in }\\ (a,b,c,d,e) = (7, 9, 13, 16, 21)\)

 

\(\left( \begin{array}{cc} \{a,b\} & 16 \\ \{a,c\} & 20 \\ \{b,c\} & 22 \\ \{a,d\} & 23 \\ \{b,d\} & 25 \\ \{a,e\} & 28 \\ \{c,d\} & 29 \\ \{b,e\} & 30 \\ \{c,e\} & 34 \\ \{d,e\} & 37 \\ \end{array} \right)\)

.
 Feb 25, 2019
 #2
avatar+104756 
+1

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

 

 

cool  cool  cool

 Feb 25, 2019

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