A group of students won $\char36{}226$ in a competition and shared the amount equally, leaving $\char36{}4$ leftover. Competing for a second time, the same group of students won $\char36{}296$. They shared the $\char36{}296$ along with the $\char36{}4$ left over from the first competition equally, with no amount remaining. Assuming that each student received a whole number of dollars in both distributions, find the greatest possible number of students in the group. Answer ...students

1. Form the divisibility conditions for the number of students

Let the number of students be $n$.

From the first competition, sharing $226 equally leaves $4, so the amount actually shared exactly is
$226-4=222.$
Hence $n$ must be a factor of $222$.

For the second distribution, they share
$296+4=300$
equally with no remainder, so $n$ must also be a factor of $300$.

2. Find the greatest possible common factor

The greatest possible value of $n$ is
$\text{HCF}(222,300).$

Using the Euclidean method,
$300-222=78,$
$222=2(78)+66,$
$78=1(66)+12,$
$66=5(12)+6,$
$12=2(6).$

So
$\text{HCF}(222,300)=6.$

However, each student must receive a whole number of dollars in the first distribution. Since $222\div 6=37$, this is valid. Also, $300\div 6=50$, valid.