Find  $x$,  where  $x$  is  the  smallest  positive  integer  such  that  $2^x$  leaves  a remainder  of  $1$  when  divided  by  $5$,  $7$,  and  $31$.

$\text{(A) } 15 \quad
\text{(B) } 20 \quad
\text{(C) } 25 \quad
\text{(D) } 30 \quad
\text{(E) } 28 \quad
\text{(F) } 32 \quad
\text{(G) } 64 \quad \\
\text{(H) } 128 \quad
\text{(I) } 45 \quad
\text{(J) } 50 \quad
\text{(K) } 60 \quad
\text{(L) } 70 \quad
\text{(M) } 80 \quad
\text{(N) } \text{none of the above}\quad $
To find the smallest positive integer \( x \) such that \( 2^x \equiv 1 \pmod{5} \), \( 2^x \equiv 1 \pmod{7} \), and \( 2^x \equiv 1 \pmod{31} \), we need to determine the order of 2 modulo 5, 7, and 31.

1. **Finding the order of 2 modulo 5:**
   \[
   \begin{aligned}
   2^1 &\equiv 2 \pmod{5}, \\
   2^2 &\equiv 4 \pmod{5}, \\
   2^3 &\equiv 3 \pmod{5}, \\
   2^4 &\equiv 1 \pmod{5}.
   \end{aligned}
   \]
   The order of 2 modulo 5 is 4.

2. **Finding the order of 2 modulo 7:**
   \[
   \begin{aligned}
   2^1 &\equiv 2 \pmod{7}, \\
   2^2 &\equiv 4 \pmod{7}, \\
   2^3 &\equiv 1 \pmod{7}.
   \end{aligned}
   \]
   The order of 2 modulo 7 is 3.

3. **Finding the order of 2 modulo 31:**
   \[
   \begin{aligned}
   2^1 &\equiv 2 \pmod{31}, \\
   2^2 &\equiv 4 \pmod{31}, \\
   2^3 &\equiv 8 \pmod{31}, \\
   2^4 &\equiv 16 \pmod{31}, \\
   2^5 &\equiv 32 \equiv 1 \pmod{31}.
   \end{aligned}
   \]
   The order of 2 modulo 31 is 5.

4. **Finding the least common multiple (LCM):**
   The smallest \( x \) must be a multiple of the orders found above. Therefore, we need to find the LCM of 4, 3, and 5.
   \[
   \text{LCM}(4, 3, 5) = 60.
   \]

Thus, the smallest positive integer \( x \) such that \( 2^x \equiv 1 \pmod{5} \), \( 2^x \equiv 1 \pmod{7} \), and \( 2^x \equiv 1 \pmod{31} \) is \( x = 60 \).

The final answer is \( \boxed{60} \).