1+3, 1+5, !+7, and !+9 (there are 4 unique sums - 4, 6,8 and 10) 3+5, 3+7, 3+9 (notice I did not pair 3 with 1 and the the only new sum is 12) 5+7, 5+9 (the only new sum is 14) 7+9 (16 is a new sum)
The sums (no repeats) are 4,6,8,10,12,14 and 16 for a total of seven numbers.
Use the distributive property to express the sum of the 2 whole numbers 15 and 30 with common factors as a multiple of two whole numbers with a sum of no common factor the solution would be (15 plus 30) = 15(1+2)