Let $(a_n)_{n=1}^{\infty}$ be an integer sequence such that $a_{n+48} \equiv a_n \pmod {35}$ for every $n \geq 1$. Let $i$ and $j$ be the least numbers satisfying the conditions $a_{n+i} \equiv a_n \pmod {5}$ and $a_{n+j} \equiv a_n \pmod {7}$ for every $n\geq 1$. Which one below cannot be an $(i,j)$ pair? 

$ 
\textbf{(A)}\ (16,4)
\qquad\textbf{(B)}\ (3,16) 
\qquad\textbf{(C)}\ (8,6)
\qquad\textbf{(D)}\ (1,48)
\qquad\textbf{(E)}\ (16,18)
$
1. Given the sequence \((a_n)_{n=1}^{\infty}\) such that \(a_{n+48} \equiv a_n \pmod{35}\) for every \(n \geq 1\), we need to find the least numbers \(i\) and \(j\) such that \(a_{n+i} \equiv a_n \pmod{5}\) and \(a_{n+j} \equiv a_n \pmod{7}\) for every \(n \geq 1\).

2. Since \(a_{n+48} \equiv a_n \pmod{35}\), it implies that \(a_{n+48} \equiv a_n \pmod{5}\) and \(a_{n+48} \equiv a_n \pmod{7}\). This means that 48 is a common period modulo 5 and modulo 7.

3. We need to find the least period \(i\) such that \(a_{n+i} \equiv a_n \pmod{5}\). The possible values of \(i\) are the divisors of 5. The least period \(i\) can be 1, 5, or any multiple of 5.

4. Similarly, we need to find the least period \(j\) such that \(a_{n+j} \equiv a_n \pmod{7}\). The possible values of \(j\) are the divisors of 7. The least period \(j\) can be 1, 7, or any multiple of 7.

5. We are given the pairs \((i, j)\) and need to determine which one cannot be a valid pair:
   - \((16, 4)\)
   - \((3, 16)\)
   - \((8, 6)\)
   - \((1, 48)\)
   - \((16, 18)\)

6. Let's analyze each pair:
   - \((16, 4)\): 16 is a multiple of 1 (valid for modulo 5), and 4 is not a multiple of 7 (invalid for modulo 7).
   - \((3, 16)\): 3 is not a multiple of 5 (invalid for modulo 5), and 16 is a multiple of 1 (valid for modulo 7).
   - \((8, 6)\): 8 is a multiple of 1 (valid for modulo 5), and 6 is not a multiple of 7 (invalid for modulo 7).
   - \((1, 48)\): 1 is a multiple of 1 (valid for modulo 5), and 48 is a multiple of 7 (valid for modulo 7).
   - \((16, 18)\): 16 is a multiple of 1 (valid for modulo 5), and 18 is not a multiple of 7 (invalid for modulo 7).

7. From the analysis, the pairs \((16, 4)\), \((3, 16)\), \((8, 6)\), and \((16, 18)\) are invalid because they do not satisfy the conditions for both moduli.

8. The pair \((1, 48)\) is valid because 1 is a multiple of 1 (valid for modulo 5), and 48 is a multiple of 7 (valid for modulo 7).

The final answer is \( \boxed{ (16, 4) } \) cannot be an \((i, j)\) pair.