Block of integers: Divisibility Let $a < b$ be natural numbers. Prove that every block of $b$ consecutive natural numbers contains two distinct elements whose product is divisible by $ab$. (I've proved this)
Suppose now $a < b < c$ are natural numbers: prove that every block of $c$ consecutive natural numbers contain three distinct elements whose product is divisible by $abc$?
Attempt to the second part: 
I'm trying to relate the case of two numbers in the first part. I can find two distinct numbers in the "$c$-block" such that their product is divisible by $ac$ or by $bc$. Problem is now getting this third distinct number, in particular I want to form the "$ab$" product. I've tried looking at a "$b$-block" within that "$c$-block" but I haven't been able to come up with a proof. 

This is my attempt.
Since the block is of size $c$, there is only one number which is divisible by $c$. Since $b<c$, there is "at least one" number (can be 1 or more) which is divisible by $b$. Again $a<b$, So for every block $b$ you choose (by dividing the $c$ block into many partitions so that each partition contains exactly one number divisible by $b$ and all the numbers in it are consecutive), so there is "at least one" number in every partition which is divisible by $a$. Thus, we can select three different numbers whose product is divisible by $abc$. 
