Let these numbers are $a,a+1,...,a+5$ It is true , that $a(a+3)+(a+1)(a+2) \leq (a+4)(a+5) \to a \leq 6 $. Easy to check $ \pmod 3$ ,so both numbers, that are divisible by $3$ must be in same product. Case $a=1$ : $6*3+1*2+4*5$ Case $a=2$: $3*6+2*5=4*7$ Case $a=3,4$: check $\pmod 4$ Case $a=5$: check $\pmod 5$ Case $a=6$ : $7*8+6*9=10*11$

sqing wrote: Determine all the sets of six consecutive positive integers such that the product of some two of them . added to the product of some other two of them is equal to the product of the remaining two numbers. It's an obvious problem.In my solution,i checked a lot of case.The answers are:{1,2,3,4,5,6},{2,3,4,5,6,7},{6,7,8,9,10,11}.

RagvaloD wrote: Let these numbers are $a,a+1,...,a+5$ It is true , that $a(a+3)+(a+1)(a+2) \leq (a+4)(a+5) \to a \leq 6 $. Easy to check $ \pmod 3$ ,so both numbers, that are divisible by $3$ must be in same product. Case $a=1$ : $6*3+1*2+4*5$ Case $a=2$: $3*6+2*5=4*7$ Case $a=3,4$: check $\pmod 4$ Case $a=5$: check $\pmod 5$ Case $a=6$ : $7*8+6*9=10*11$ Why does this inequality hold ?

sqing wrote: Determine all the sets of six consecutive positive integers such that the product of some two of them . added to the product of some other two of them is equal to the product of the remaining two numbers.

Let the numbers be $a,a+1,a+2,a+3,a+4,a+5$. If $a\geq 7$, we have $(a+4)(a+5)< a(a+3)+(a+1)(a+2)$, contradiction since the left side is the largest possible value of a product and the right side is the smallest possible value of the sum of 2 products. Bashing all pairings (there are 15 for each $1\leq a\leq 6$) gives us that we have $a=1,2,6$, so our answer is the sets of consecutive positive integers starting with either 1, 2, or 6.