117*. The square of a prime number, different from $2,3,5$, when divided by 30, gives a remainder equal to 1 or 19.
117. \( p^{2} = 30m + r \), where \( r < 30 \). Considering that \( p \) is not divisible by \( 2, 3, 5 \), and that the square of an odd number ends only in the digits \( 1, 5, 9 \), we conclude that \( r \) can only be \( 1, 11, 19, 29 \). However, \( r \neq 11 \), because \( p^{2} - 11 = p^{2} - 1 - 10 \) is not divisible by 3 according to Fermat's theorem, hence \( p^{2} - 11 \) is not divisible by 30.

Similarly, \( p^{2} - 29 = p^{2} - 1 - 28 \) is not divisible by 30, hence \( r \neq 29 \). We conclude that \( r = 1; 19 \).