Answer:
14 students
Step-by-step explanation:
Let x be the number of students, [tex]x<70.[/tex]
From the first congruence,
[tex]x=2+4k,[/tex]
substitute it into the second congruence:
[tex]2+4k\equiv 1(\rm mod\ 5)\\ \\4k\equiv -1(\rm mod\ 5)\\ \\4k\equiv -1+5(\rm mod \ 5)\\ \\4k\equiv 4(\rm mod \ 5)\\ \\k\equiv 1(\rm mod \ 5)\\ \\k=5l+1\\ \\x=4(5l+1)+2=20l+6[/tex]
Substitute this into the third congruence:
[tex]20l+6\equiv 4(\rm mod\ 6)\\ \\20l\equiv -2(\rm \ mod 6)\\ \\20l\equiv -2+6\cdot 7(\rm mod\ 6)\\ \\20l\equiv 40(\rm mod \ 6)\\ \\l\equiv 2(\rm mod 6)\\ \\l=6m+2\\ \\x=20(6m+2)+6=120m+46[/tex]
Since [tex]x<70,[/tex] then [tex]x=46\ (\text{when }m=0)[/tex]
The next number divisible by 4, 5 and 6, greater than 46 and less than 70 is 60, so 14 students should be added.