I3.4 The total number of students in a school is a multiple of 7 and not less than 1000 . Given that the same remainder 1 will be obtained when the number of students is divided by $p+1, p+2$ and $p+3$. Let $q$ be the least of the possible numbers of students in the school, find $q$.
$\begin{array}{l}p+1=3, p+2=4, p+3=5 ; \mathrm{HCF}=1, \mathrm{LCM}=60 \\ q=60 m+1, \text { where } m \text { is an integer. } \\ \because q \geq 1000 \text { and } q=7 k, k \text { is an integer. } \\ 60 m+1=7 k \\ 7 k-60 m=1 \\ k=43, m=5 \text { satisfies the equation } \\ k=43+60 t ; 7 k \geq 1000 \Rightarrow 7(43+60 t) \geq 1000 \Rightarrow t \geq 2 \Rightarrow \text { Least } q=7 \times(43+60 \times 2)=1141\end{array}$