= Solution
A <linear congruential generator> starts with an integer seed $S_0\in\{0,\ldots,M-1\}$ and iterates
$$
S_{i+1}=(aS_i+c)\bmod M,\qquad U_i=S_i/M.
$$
The outputs lie in $[0,1)$ and are used as <pseudorandom numbers> approximating a <uniform distribution>. They are deterministic once the seed is fixed. The largest possible period is $M$, since there are only $M$ states; good period alone does not establish <independence> or good multivariate uniformity.
The <full-period criterion for a mixed congruential generator> says that period $M$ for every seed occurs exactly when
$$
\boxed{\gcd(c,M)=1,\quad p\mid(a-1)\text{ for every prime }p\mid M,\quad
4\mid M\ \Longrightarrow\ 4\mid(a-1).}
$$
For $M=12^k=2^{2k}3^k$, these conditions reduce to
$$
\boxed{a\equiv1\pmod{12},\qquad \gcd(c,6)=1.}
$$
The multiplier must satisfy both $a\equiv1\pmod3$ and $a\equiv1\pmod4$. A positive shift alone is insufficient: for example, a shift divisible by three cannot give period $M$ when the multiplier is one modulo three. If the required coprime shift has already been chosen, the condition on the multiplier alone is $a=1+12r$ for an integer $r$.
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