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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.

 

Thanks.

 Mar 17, 2021
 #1
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help!!

 Mar 17, 2021
edited by MathyGoo13  Mar 17, 2021
 #2
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Let the numbers be a < b < c < d < e

Then the smallest two, a and b, must have sum 16,

a + b = 16

and the largest two, d and e, must have sum 37,

d + e = 37.

Then the middle number c is equal to 66 - (a+b) - (d+e) = 66 - 16 - 37 = 13.

 OK. 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.

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.

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 is e+c.

It can not be nothing else.

If e + c = 34 and c = 13, then e = 34 - 13 = 21.

It implies d = 37-21 = 16.

Thus the numbers are 7, 9, 13, 16 and 21.

Their product is 275184.

 

hope this helps!

 Mar 17, 2021
 #3
avatar+130 
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Not to be rude, but thanks for the solution off of Algebra.com. 

I need to know where we got 66 from!

 Mar 17, 2021
 #4
avatar+31 
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37 + 29

this is quite an interesting problem indeed, I apologize for not explaining, it's new to me as well!

owentout  Mar 17, 2021
 #5
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wait! I'm a doofus, add all the given numbers and divide by 4 ( 4 because you are taking into account the numbers excluding the one you are looking at)

THAT would get you 66

 Mar 17, 2021
 #6
avatar+130 
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Oh thanks for the answer! I wasn't expecting a reply! Thanks! I get it now!

MathyGoo13  Mar 18, 2021
 #7
avatar+130 
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Wait, why are we taking into account all the numbers except for the one we are looking at? Can you tell me?

MathyGoo13  Mar 18, 2021

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