Question

The Mathclub, VIT-AP wants to conduct a group event for its members. So the club president has to fix the group size the event. When he tries to fix the size to be 5 members in each group, 4 members are left; when he tries to fix the size to be 6 members in each group, 5 members are left; When he fixes the size to be 7 members in each group, 6 members are left. What is the smallest number of members that the club has?

Answer 1

Let $x$ be the number of members that the club has. Then we have the following system of linear congruences:

$\begin{cases} x\equiv 4\ (\mod 5)\\ x\equiv 5\ (\mod 6)\\ x\equiv 6\ (\mod 7) \end{cases}$

The first congruence is equivalent to the equality $x=4+5t,\ t\in\mathbb Z.$ Put this in the second congruence:

$4+5t\equiv 5\ (\mod 6)$

$5t\equiv 1\ (\mod 6)$

$-t\equiv 1\ (\mod 6)$

$t\equiv -1\ (\mod 6)$

Consequently, $t=-1+6s,\ s\in\mathbb Z.$

$x=4+5t=4+5(-1+6s)=-1+30s$

Put $x$ in the third congruence of the system:

$-1+30s\equiv 6\ (\mod 7)$

$30s\equiv 7\ (\mod 7)$

$30s\equiv 0\ (\mod 7)$

$s\equiv 0\ (\mod 7)$

Therefore, $s=7k,\ k\in\mathbb Z.$

We conclude that

$x=-1+30s=-1+30\cdot 7k=-1+210k, k\in\mathbb Z.$

For we have the smallest positive $x=-1+210=209$.

Therefore, the smallest number of members that the club has is 209.