Question 1027853
<pre>
We 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.

[Those are the only two pairs that we can be sure
have the smallest and largest sums.  We can't be
sure about the other pairs.]

If the smallest number, a, is too small we won't
be able to get all those large sums.  So let's
take the smallest, a, to be as large as possible 
We'll begin by trying a=7 and b=9, so that 7+9=16.

We need to be able to get 20 by adding two numbers, 
so we need to add 13 to the 7 to get the 20. [Note
that we'd have to add 4 to the 9, and 4 is too small.] 

We take c=13, so that 7+13=20.

We can now also get 22, since 9+13=22.

We have so far, a=7, b=9, c=13. 

So d is larger than 13.

d can't be 14 because 14+13=27 which is not one of 
the sums.

d can't be 15 because 15+9=24 which is not one of 
the sums.

But d can be 16, because 7+16=23, which is one of 
the sums.
   
Also 9+16=25, which is one of the sums.

We have so far, a=7, b=9, c=13, d=16

Now since d + e = 37, e can only be 21, 

so that 16+21=37

Answer: a=7, b=9, c=13, d=16, e=21

Checking:

 7+ 9 = 16
 7+13 = 20
 9+13 = 22
 7+16 = 23
 9+16 = 25
 7+21 = 28
13+16 = 29
 9+21 = 30
13+21 = 34
16+21 = 37

The product is 7*9*13*16*21 = 275184

Edwin</pre>