We need to find the smallest number of pennies $x$ that satisfies the following conditions:

1. When divided by 2, there is 1 penny left over.
   $x \equiv 1 \pmod{2}$

2. When divided by 3, there is 1 penny left over.
   $x \equiv 1 \pmod{3}$

3. When divided by 4, there is 1 penny left over.
   $x \equiv 1 \pmod{4}$

4. When divided by 5, there is 1 penny left over.
   $x \equiv 1 \pmod{5}$

5. When divided by 6, there is 1 penny left over.
   $x \equiv 1 \pmod{6}$

6. When divided by 7, there is no remainder.
   $x \equiv 0 \pmod{7}$

Approach:

Since the first five conditions imply $x \equiv 1 \pmod{2, 3, 4, 5, 6}$, the number $x-1$ must be divisible by the least common multiple (LCM) of 2, 3, 4, 5, and 6. The LCM of these numbers is 60.

So, $x-1$ must be a multiple of 60, meaning: $x = 60k + 1$ for some integer $k$. Additionally, $x$ must also be divisible by 7: $60k + 1 \equiv 0 \pmod{7}$ $60k \equiv -1 \pmod{7}$ Since $60 \equiv 4 \pmod{7}$, this simplifies to: $4k \equiv -1 \pmod{7}$ Multiplying both sides by the modular inverse of 4 modulo 7 (which is 2), we get: $8k \equiv -2 \pmod{7}$ $k \equiv 5 \pmod{7}$

Thus, $k = 7m + 5$ for some integer $m$, and substituting into $x = 60k + 1$: $x = 60(7m + 5) + 1 = 420m + 301$

The smallest solution is when $m = 0$, which gives: $x = 301$

Thus, the smallest number of pennies the student could have is 301.

Would you like further details or have any questions? Here are some follow-up questions for deeper understanding:

1. How is the least common multiple (LCM) of several numbers calculated?
2. What is the significance of modular arithmetic in problem-solving?
3. How can we find the modular inverse of a number?
4. What other problems can be solved using the Chinese Remainder Theorem?
5. Can this problem be solved using a different method?

Tip: When solving problems involving multiple remainders, the Chinese Remainder Theorem is often a helpful tool to find solutions systematically.