pohoatza wrote: Positive integers $ a<b$ are given. Prove that among every $ b$ consecutive positive integers there are two numbers whose product is divisible by $ ab$. Let $ S = \{n+1, n+2, ..., n+b\}$. Clearly there is a single $ (n+i)$ in $ S$ that is divisible by $ b$ because the remainders of all elements of $ S$ divided by $ b$ must belong to the set $ \{0, 1, ..., b-1\}$. Taking $ a$ consecutive integers in $ S$ we get some $ (n+j)$ divisible by $ a$. If $ i \neq j$ then we are done. If $ i = j$ then it means that $ b$ is a multiple of $ a$ because $ a<b$. So let $ b = ka$ for some $ k$. We can now partition $ S$ into $ k$ blocks so that each block has some element $ (n+j)$ divisible by $ a$. Just pick one of these $ (n+j)$ different from the $ (n+i)$ and we are done.

Why $ a|b$ ? I can't understand this. $ a,b|n+i$ just imply $ lcm(a,b)|n+i$...

Yeah, it's wrong. However, that mistake is very easy to fix. You just have to find another number divisible by $ \gcd (a,b)$

And it's obvious it exists since $ gcd(a, b) \leq \frac{b}{2}$.

O.K. I'm going to complete everyone's idea. Obviously among $b$ consecutive positive integers there are two numbers $p$ which is divisible by $a$ and $q$ which is divisible by $b$. Case1:$p\neq q$ Then $ab|pq$,and the proof is completed. Case2: $p=q$ Then $lcm(a,b)|p$.Since $gcd(a,b)\le \frac{1}{2}b$,there are at least two numbers which are divisible by $gcd(a,b)$.Then ∃$x\neq p$ s.t. $gcd(a,b)|x$.Therefore $ab=lcm(a,b)gcd(a,b)|px$,and the proof is completed. From case$1,2$,completing the proof.$\blacksquare$