G10.4A positive integer $d$ when divided by 7 will have 1 as its remainder; when divided by 5 will have 2 as its remainder and when divided by 3 will have 2 as its remainder. Find the least possible value of $d$.
$$
\begin{array}{l}
d=7 m+1=5 n+2=3 r+2 \\
7 m-5 n=1 \ldots \ldots .(1) \\
5 n=3 r \quad \ldots \ldots .(2)
\end{array}
$$

From (2), $n=3 k, r=5 k$
Sub. $n=3 k$ into $(1), 7 m-5(3 k)=1$
$\Rightarrow 7 m-15 k=1$
$-14+15=1 \Rightarrow$ A possible solution is $m=-2, k=-1$
$$
m=-2+15 t, k=-1+7 t
$$

When $t=1, m=13, k=6, n=18, r=30$.
The least possible value of $d=3(30)+2=92$