Solution :
A). x = 2 (mod 3) [tex]$\mu = 3\times 5 \times 7 = 105$[/tex]
x = 3 (mod 5) [tex]$y_1=35^{-1} (\mod 3)$[/tex]
x = 4 (mod 7) [tex]y_1=2[/tex]
[tex]$y_2=21^{-1}(\mod5) = 1$[/tex]
[tex]$y_3=15^{-1}(\mod7) = 1$[/tex]
[tex]$x=2 \times 35 \times 2 + 3\times 21\times 1+4\times 15\times 1$[/tex]
[tex]=140+63+60[/tex]
[tex]=263[/tex]
≡ 53(mod 105)
Hence the solution is 105 k + 53 > 1000 for k = 10
Therefore, the minimum number of students = 1103
B). [tex]$\phi (935) = 640$[/tex]
By Eulu's theory
[tex]$935 | a^{640}_n -1$[/tex] if n and 935 are coprime.
Now, [tex]$935|n^{80}-1$[/tex] and 80 x 8 = 640
[tex]$935|n^{640}-1$[/tex] ⇒ g(n,935) = 1
⇒ 5, 11, 17 do not divide n
