Divide the set of integers from 1 to 15 into 2 subsets - a subset A of 13 numbers and B of 2 numbers so that the sum of the members of A equals the product of the two numbers in B.

The sum of all 15 integers is 120 less the smallest two, 1 + 2, equaling 3, and the largest two, 14 + 15, equaling 29. Thus, the subset A, the sum of 13 integers of the set of integers 1 to 15, must be in the inclusive range 91 to 117.  There are only ten pairs of distinct integers in the set that will provide a product in the given range, none of which meets the product-sum condition:
{ 7×13} =  91; 120 - { 7+13} = 100;
{ 8×12} =  96; 120 - { 8+12} = 100;
{ 7×14} =  98; 120 - { 7+14} =  99;
{ 9×11} =  99; 120 - { 9+11} = 100;
{ 8×13} = 104; 120 - { 7+13} =  99;
{ 7×15} = 105; 120 - { 7+15} =  98;
{ 9×12} = 108; 120 - { 9+12} =  99;
{10×11} = 110; 120 - {10+11} =  99;
{ 8×14} = 112; 120 - { 8+14} =  98;
{ 9×13} = 117; 120 - { 9+13} =  98;

If the condition were modified into the subsets of A, the sum of 12 of the 15 numbers, and B, the product of the remaining 3 numbers, then a solution does exist:
subset A {2,3,4,5,6,7,8,10,12,13,14,15}
         [2+3+4+5+6+7+8+10+12+13+14+15 = 99]
subset B {1,9,11}
         [1×9×11 = 99]

(No claim as to being unique, and in fact has been found that two other solutions exist for the subsets of 12 and 3 elements).

Edited on March 1, 2016, 9:29 am

Posted by Dej Mar on 2016-03-01 09:17:00