Determine all sets of six consecutive positive integers such that the product of two of them, added to the product of other two of them is equal to the product of the remaining two numbers.
Exactly two of the six numbers are multiples of 3 and these two need to be multiplied together, otherwise two of the three terms of the equality are multiples of 3 but the third one is not.

Let $n$ and $n+3$ denote these multiples of 3. Two of the four remaining numbers give remainder 1 when divided by 3, while the other two give remainder 2, so the two other products are either

$\equiv 1 \cdot 1=1(\bmod 3)$ and $\equiv 2 \cdot 2 \equiv 1(\bmod 3)$, or they are both $\equiv 1 \cdot 2 \equiv 2(\bmod 3)$. In conclusion, the term $n(n+3)$ needs to be on the right hand side of the equality.

Looking at parity, three of the numbers are odd, and three are even. One of $n$ and $n+3$ is odd, the other even, so exactly two of the other numbers are odd. As $n(n+3)$ is even, the two remaining odd numbers need to appear in different terms.

We distinguish the following cases:

I. The numbers are $n-2, n-1, n, n+1, n+2, n+3$.

The product of the two numbers on the RHS needs to be larger than $n(n+3)$. The only possibility is $(n-2)(n-1)+n(n+3)=(n+1)(n+2)$ which leads to $n=3$. Indeed, $1 \cdot 2+3 \cdot 6=4 \cdot 5$

II. The numbers are $n-1, n, n+1, n+2, n+3, n+4$.

As $(n+4)(n-1)+n(n+3)=(n+1)(n+2)$ has no solutions, $n+4$ needs to be on the RHS, multiplied with a number having a different parity, so $n-1$ or $n+1$.

$(n+2)(n-1)+n(n+3)=(n+1)(n+4)$ leads to $n=3$. Indeed, $2 \cdot 5+3 \cdot 6=4 \cdot 7$. $(n+2)(n+1)+n(n+3)=(n-1)(n+4)$ has no solution.

III. The numbers are $n, n+1, n+2, n+3, n+4, n+5$.

We need to consider the following situations:

$(n+1)(n+2)+n(n+3)=(n+4)(n+5)$ which leads to $n=6$; indeed $7 \cdot 8+6 \cdot 9=10 \cdot 11$;

$(n+2)(n+5)+n(n+3)=(n+1)(n+4)$ obviously without solutions, and

$(n+1)(n+4)+n(n+3)=(n+2)(n+5)$ which leads to $n=2$ (not a multiple of 3 ).

In conclusion, the problem has three solutions:

$$
1 \cdot 2+3 \cdot 6=4 \cdot 5, \quad 2 \cdot 5+3 \cdot 6=4 \cdot 7, \quad \text { and } \quad 7 \cdot 8+6 \cdot 9=10 \cdot 11
$$