tth Coordinator

2 .We can't fine the smallest value of the expression.

1.

Because p and q are prime number > 3

=> q can equal 3k+1 or 3k+2

But if q = 3k+1 => p = 3k+1 + 2 = 3k+3 divided by 3 -> wrong

And if q = 3k+2 => p = 3k+2 + 2 = 3k+4 divided by 3 remainder 1   -> right

So p+q = 3k+2 + 3k+4 = 6k+6 = \(6\left(k+1\right)⋮6\)

We have 2 cases:

\(\left[{}\begin{matrix}k=2h+1\\k=2k\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}6\left(k+1\right)=6\cdot\left(2h+2\right)=12h+12⋮12\\6\left(k+1\right)=6\left(2h+1\right)=12h+6\equiv6\left(mod12\right)\end{matrix}\right.\)So p+q divided by 12 remainder 0 or 6