329. The number $a^{100}$ when divided by 73 gives a remainder of 2, and the number $a^{101}$ when divided by the same number gives a remainder of 69. Find the remainder of the number $a$ when divided by 73.
329. $a^{100}=73 m+2$, from which $a^{101}=73 n+2 a$ (1). In addition, $a^{101}=73 q+69(2)$. From (1) and (2) we find $73 k+2 a=$ $=69(3)$.

Noting that equation (3) is satisfied by the values $k=1$ and $a=-2$, we get $a=73 t-2$, which can be written as: $a=73 t+71$. The required remainder is 71.