Six products $(ab,ac,ad,bc,bd,cd)$, one product is removed. Remaining products are consecutive natural numbers. Deleted product is...? Step 1: Alice writes four distinct real positive numbers on a school board, say $a,b,c,d$. For example, the numbers written on the board could be $4.56$, $2$, $\sqrt{3}$ and $\pi^e$.
Step 2: After that she creates all possible pairs of these four numbers and writes their products on the board ($ab,ac,ad,bc,bd,cd$). 
Step 3: She picks one of these six numbers and erases it from the board, leaving five products on the board.
After that Alice notices that the remaining five numbers reperesent a sequence of consecutive natural numbers (e.g. $20, 21, 22, 23, 24$). 
What is the number deleted in step 3?
I started by assuming that $a<b<c<d$. It means that either $ab<ac<ad<bc<bd<cd$ or $ab<ac<bc<ad<bd<cd$. I was able to solve the problem in a "brute force" style by considering both cases and by deleting a different product on every try. Many cases lead to contradicitons (like $a=b$) but some of them always lead to the same result (I'll keep it secret for the time being). But I don't like the solution: it is quite messy and I would have to type a lot if I wanted to present it here. 
I am thinking there has to be a solution which is short and simple but I was not able to find it.
 A: There are three pairs among the products that each have the same product - $ab\cdot cd$, $ac\cdot bd$, $ad\cdot bc$. After removing one of the products, two pairs remain, as well as one unpaired product. Suppose WLOG we removed $cd$.
Now, either both $ac$ and $bd$ are between $ad$ and $bc$, or both $ad$ and $bc$ are between $ac$ and $bd$. WLOG, it's the former. Then, since they're all integers, $|ad-bc|\ge 3$. If $|ad-bc|=3$, then since $n(n+3)=(n+1)(n+2)-2$ for all $n$, the products aren't equal. That leaves $|ad-bc|=4$; let $n$ be whichever of $ad$ and $bc$ is smaller. Since we have $n(n+4)=(n+1)(n+3)-3$, $ac\cdot bd < (n+1)(n+3)$ and the products $ac,bd$ must be $n+1$ and $n+2$ in some order.
Now, $n(n+4)=(n+1)(n+2)$, so $n^2+4n=n^2+3n+2$ and $n=2$. Our pairs are $(2,6)$ and $(3,4)$, leaving $5$ unpaired. The removed number $bd$ must form a pair of product $12$ with $5$, so it was $\frac{12}{5}=\boxed{2.4}$. Done.
Well, almost done. We've established what the removed number had to be, but we haven't established that it was possible at all. That just requires an example. Playing with it a bit, I find $a=2\sqrt{\frac56}$, $b=3\sqrt{\frac56}$, $c=2\sqrt{\frac65}$, and $d=\sqrt{\frac65}$ works.
