Omkar, $\mathrm{Krit}_{1}, \mathrm{Krit}_{2}$, and $\mathrm{Krit}_{3}$ are sharing $x>0$ pints of soup for dinner. Omkar always takes 1 pint of soup (unless the amount left is less than one pint, in which case he simply takes all the remaining soup). Krit $_{1}$ always takes $\frac{1}{6}$ of what is left, Krit $_{2}$ always takes $\frac{1}{5}$ of what is left, and $\mathrm{Krit}_{3}$ always takes $\frac{1}{4}$ of what is left. They take soup in the order of Omkar, $\mathrm{Krit}_{1}, \mathrm{Krit}_{2}, \mathrm{Krit}_{3}$, and then cycle through this order until no soup remains. Find all $x$ for which everyone gets the same amount of soup.
The main observation is that if $x>1$ pints of soup are left, then in one round, Omkar gets 1 and each Krit $_{n}$ gets $\frac{x-1}{6}$, with $\frac{x-1}{2}$ soup left. Thus it is evident that each Krit $_{n}$ gets the same amount of soup, which means it suffices to find $x$ for which Omkar gets $\frac{x}{4}$. Omkar gets 1 for each cycle and then all the remaining soup when there is less than one pint remaining. The amount of soup becomes (after each cycle) \[ x \rightarrow \frac{x-1}{2} \rightarrow \frac{x-3}{4} \rightarrow \cdots \rightarrow \frac{x+1}{2^{n}}-1 \] so if $n$ is the number of cycles, then Omkar's soup is $n+\frac{x+1}{2^{n}}-1$. Setting this equal to $\frac{x}{4}$, we obtain \[ x=\frac{n+1 / 2^{n}-1}{1 / 4-1 / 2^{n}}=\frac{(n-1) 2^{n}+1}{2^{n-2}-1} \] This immediately implies $n>2$. On the other hand, we necessarily have $0 \leq \frac{x+1}{2^{n}}-1 \leq 1$, so $2^{n} \leq x+1 \leq 2^{n+1}$. But \[ x+1=\frac{(n-1) 2^{n}+2^{n-2}}{2^{n-2}-1} \leq \frac{(n-1) 2^{n}+2^{n}}{2^{n-3}}=8 n \] So $2^{n} \leq 8 n \Longrightarrow n \leq 5$. Testing $n=3,4,5$ : \begin{itemize}   \item For $n=3$ we get $x=17$ which is greater than $2^{4}$.   \item For $n=4$ we get $x=\frac{49}{3}$ which works.   \item For $n=5$ we get $x=\frac{129}{7}$ which is less than $2^{5}$. \end{itemize} We see that only $n=4$ and $x=\frac{49}{3}$ works. $\fbox{\frac{49}{3}}$.