Grace made over four dozen cupcakes. If she makes packages of 2 cupcakes, then there is 1 left over. If she makes packages of 3 cupcakes, then there are 2 left over. If she makes packages of 4 cupcakes, then there are 3 left over. What is the fewest number of cupcakes she could have made?

To solve this problem, we are given that Grace made over four dozen cupcakes and are asked to find the fewest number of cupcakes she could have made based on the following conditions:

- If she makes packages of 2 cupcakes, there is 1 left over.
- If she makes packages of 3 cupcakes, there are 2 left over.
- If she makes packages of 4 cupcakes, there are 3 left over.

Let's define the number of cupcakes Grace made as \( x \).

### Step 1: Set up the congruences
From the problem, we can form the following congruences:
1. \( x \equiv 1 \pmod{2} \)
2. \( x \equiv 2 \pmod{3} \)
3. \( x \equiv 3 \pmod{4} \)

We are also told that Grace made more than four dozen cupcakes, meaning \( x > 48 \).

### Step 2: Solve the system of congruences

To find the smallest \( x \), we will check numbers that satisfy all three conditions.

Starting from \( x = 3 \) (as \( x \equiv 3 \pmod{4} \)):

- \( x = 3 \): does not satisfy \( x \equiv 2 \pmod{3} \).
- \( x = 7 \): does not satisfy \( x \equiv 2 \pmod{3} \).
- \( x = 11 \): does not satisfy \( x \equiv 2 \pmod{3} \).
- \( x = 15 \): does not satisfy \( x \equiv 2 \pmod{3} \).
- \( x = 19 \): does not satisfy \( x \equiv 2 \pmod{3} \).
- \( x = 23 \): does not satisfy \( x \equiv 2 \pmod{3} \).
- \( x = 27 \): satisfies both \( x \equiv 2 \pmod{3} \) and \( x \equiv 1 \pmod{2} \).

The smallest \( x \) that satisfies all three conditions is 27, but Grace made over 48 cupcakes.

### Step 3: Adjust for the constraint \( x > 48 \)

Since \( x > 48 \), we now continue finding the next values that satisfy all three congruences:

Continuing from \( x = 27 \), we add 12 (the least common multiple of 2, 3, and 4) to get the next possible solutions:
- \( x = 27 + 12 = 39 \)
- \( x = 39 + 12 = 51 \)

Thus, the smallest number of cupcakes that satisfies all the conditions and is greater than 48 is \( \boxed{51} \).

Let me know if you would like further clarification!

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Would you like more details or have any questions about this solution? Here are five related questions to expand your understanding:

1. What is the Chinese Remainder Theorem, and how could it be used to solve this type of problem?
2. How would the solution change if the conditions involved packages of 5 or 6 cupcakes instead?
3. How can we systematically approach modular arithmetic problems involving multiple conditions?
4. Why is it necessary to account for the constraint that Grace made more than four dozen cupcakes?
5. What are the advantages of solving modular arithmetic problems by hand versus using a systematic algorithm?

**Tip:** When solving problems with remainders (modular arithmetic), break them down into simple congruences and then look for common solutions among them.

Discover how to find the fewest number of cupcakes Grace made using modular arithmetic. This problem involves solving congruences such as x ≡ 1 (mod 2), x ≡ 2 (mod 3), and x ≡ 3 (mod 4), applying the least common multiple concept. Suitable for students in Grades 9-11.

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