A student has a penny collection. When the student puts the Pennie’s in groups of 2 there is one penny left over. When they are put in groups of three, four, five, and six there is also one penny left over. When they are put in groups of seven there are not Pennie’s left over. What is the smallest number of pennies the student could have?

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.

A student has a penny collection. When the student puts the pennies in groups of 2 there is one penny left over. When they are put in groups of three, four, five, and six there is also one penny left over. When they are put in groups of seven there are no pennies left over. What is the smallest number of pennies the student could have?

Find the smallest number of pennies that satisfy multiple modular conditions, including divisions by 2, 3, 4, 5, and 6 with remainders, and by 7 with no remainder. This problem explores key concepts like the Least Common Multiple (LCM) and the Chinese Remainder Theorem, suitable for high school students.

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