Distinct Pairwise Sums

Since the pairwise sums are distinct, no two of the numbers can be the same. Let the 4 numbers be $a<b<c<d$ . The 6 sums will be $a+b, a+c, a+d, b+c, b+d$ , and $c+d$ . We can easily see that $c+d$ and $b+d$ are the 2 largest sums, with $b+d < c+d$ . We can also easily see that $a+b$ and $a+c$ are the 2 smallest sums, with $a+b<a+c$ . Thus $a+b=7, a+c=9$ , so $c-b=2$ and note there are 2 cases to deal with.

Case 1: $a+d=10, b+c=14$ , With $c-b=2$ , this implies that $c=8, b=6$ . Using $a+c=9$ , we get $a=1$ . Using $a+d=10$ , we get $d=9$ , thus $abcd=432$ .

Case 2: $b+c=10, a+d=14$ . With $c-b=2$ , this implies that $c=6, b=4$ . Using $a+c=9$ , we get $a=3$ . Using $a+d=14$ , we get $d=11$ , thus $abcd=792$ .

Thus, the largest possible product is 792.

Let the 4 numbers be $a, b, c$ and $d$ . Since the pairwise sums are distinct, the 4 numbers are distinct. Without loss of generality, let $a < b < c < d$ . Observe that $a+b$ is the smallest pairwise sum, and $a + c$ is the second smallest pairwise sum. Hence we have $a+b= 7$ and $a+c = 9$ , thus $a= \frac{16-(b+c)}{2}$ . Likewise, $d+c$ is the largest pairwise sum, and $b + d$ is the second largest pairwise sum. Now we have the following cases:

Case 1: $a+d = 10$ and $b+c = 14$ . So $a = \frac{16-14}{2} = 1$ , and substituting this into the other three gives $b = 6, c =8$ and $d = 9$ . Thus the product is $1\times 6\times 8\times 9 = 432$ .

Case 2: $a+d = 14$ and $b+c = 10$ . So $a = \frac{16-10}{2} = 3$ , and substituting this into the other three gives $b = 4, c =6$ and $d = 11$ . Thus the product is $3\times 4\times 6\times 11 = 792$ .

Hence, the largest possible product is $792$ .

I solved some numbers from 1 to 13 as in quest max no is 14 so last possibility of sum of 14 in my mind is that one so think then the 4 numbers b/w 1 and 13 and their check their sum as smallest sums are 7,9,10 and 14. Now the method 3 + 4 = 7, 4+6 =10, 3+6 = 9 , 3+11 = 14, and other largest two sums are 4+11=15 and 6+11 = 17 where 3 4 6*11=792

a < b < c < d

a + b = 6

largest a * b=12 with a = 3 and b = 4

a + c = 9 --> c = 9 - 3 = 6

b + c = 10 (correct)

a + d = 14 --> d = 14 - 3 = 11

so: a = 3 ; b = 4 ; c = 6 ; d = 11

then a * b * c * d = 792 (largest).

Suppose the 4 numbers are A, B, C, D. Since the sums are distinct, the numbers must also be distinct. For the product to be as large as possible, the four numbers need to be as close as possible. Without loss of generality, suppose A<B<C<D, we get A+B=7, A+C=9, A+D=14, B+C=10 (for A,B,C,D to be as close as possible). We then get 2(A+B+C)=26, so A+B+C=13 We get A=3, B=4, C=6, D=11, so the product is 792