一组学生在一次比赛中赢得了 $\char36{}226$，并平分了这笔奖金，最后剩下 $\char36{}4$。在第二次比赛中，同一组学生赢得了 $\char36{}296$。他们平分了这 $\char36{}296$ 以及第一次比赛剩下的 $\char36{}4$，没有剩下任何钱。 假设在两次分配中每个学生分到的金额都是整数元，求这组学生的最大可能人数。 答案 ...名学生

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.