Question 1027853
.
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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


I agree with Mr. Edwin McCravy that my solution was wrong.
Mr. Edwin McCravy, thank you for your critics and detecting the error.


I still keep my old solution below for documentary purposes (but probably will delete it later).

Instead, I will put here the other solution, which is shorter.


<pre>
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.

   ----------------------------------------------------------------------
   Previous 5 lines are exactly from Mr. McCravy solution, but what follows is different.
   ----------------------------------------------------------------------

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

(The number 66 goes from my old solution which is correct to this point).

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 (see the condition) 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.
</pre>

Below is my old &nbsp;>>>wrong<<< &nbsp;solution.
It is still valid in the part that the sum of the five numbers is 66.
What follows after that is wrong.
<pre>
Let  {{{a[1]}}},  {{{a[2]}}},  {{{a[3]}}},  {{{a[4]}}}  and  {{{a[5]}}}  be our five different positive integers.

We are given that

{{{a[1]}}} + {{{a[2]}}} = 16,     (1)
{{{a[1]}}} + {{{a[3]}}} = 20,     (2)
{{{a[1]}}} + {{{a[4]}}} = 22,     (3)
{{{a[1]}}} + {{{a[5]}}} = 23,     (4)
{{{a[2]}}} + {{{a[3]}}} = 25,     (5)
{{{a[2]}}} + {{{a[4]}}} = 28,     (6)
{{{a[2]}}} + {{{a[5]}}} = 29,     (7)
{{{a[3]}}} + {{{a[4]}}} = 30,     (8)
{{{a[3]}}} + {{{a[5]}}} = 34,     (9)
{{{a[4]}}} + {{{a[5]}}} = 37.     (10)

Add all ten equations (1) - (10) at once (both left and right sides). You will get an equation

{{{4*(a[1] + a[2] + a[3] + a[4] + a[5])}}} = {{{264}}},    or,  after dividing both sides by 4

{{{a[1] + a[2] + a[3] + a[4] + a[5]}}} = {{{66}}}.    (*)

Now distract, one by one and one after one, the above equations (10), (9), (8), (7) and (3) from the equation (*) as it is shown below. 

{{{a[1] + a[2] + a[3]}}} = 29,     (11)    ( distract (10) from (*);   29 = 66 - 37 )
{{{a[1] + a[2] + a[4]}}} = 32,     (12)    ( distract  (9) from (*);   32 = 66 - 34 )
{{{a[1] + a[2] + a[5]}}} = 36,     (13)    ( distract  (8) from (*);   36 = 66 - 30 )
{{{a[1] + a[3] + a[4]}}} = 37,     (14)    ( distract  (7) from (*);   37 = 66 - 29 )
{{{a[2] + a[3] + a[5]}}} = 44,     (15)    ( distract  (3) from (*);   44 = 66 - 22 )

Now,  compare (1) and (11). You will get  {{{a[3]}}} = 29 - 16 = 13.   (16)
Next, compare (1) and (12). You will get  {{{a[4]}}} = 32 - 16 = 16.   (17)
      Compare (1) and (13). You will get  {{{a[5]}}} = 36 - 16 = 20.   (18)
      Compare (8) and (14). You will get  {{{a[1]}}} = 37 - 30 =  7.   (19)
      Compare (9) and (15). You will get  {{{a[2]}}} = 44 - 34 = 10.   (20)

Now, to answer the question, multiply the right sides of (16) - (20). You will get

{{{a[1]*a[2]*a[3]*a[4]*a[5]}}} = 13*16*20*7*10 = 291200.

<U>Answer</U>.  The product of five numbers is 291200.
</pre>