Kids math problem.
A chocolate shop is closing down and needs to sell all their chocolates. Your Mum only allows you to buy enough to fit in her handbag.
The chocolates were counted in groups of 10 but there were 3 left over.
The chocolates were counted in groups of 2 but there was 1 left over.
How many chocolates could you buy? How many possible answers can you come up with?

This problem involves finding a number of chocolates that fits the conditions when grouped in 10s and 2s. Let’s break this down.

### Problem Statement:
- **Groups of 10:** There are 3 chocolates left over when counted in groups of 10. Mathematically, this is written as:
  \[
  N \equiv 3 \ (\text{mod} \ 10)
  \]
  This means the total number of chocolates \( N \) gives a remainder of 3 when divided by 10.

- **Groups of 2:** There is 1 chocolate left over when counted in groups of 2. Mathematically, this is written as:
  \[
  N \equiv 1 \ (\text{mod} \ 2)
  \]
  This means \( N \) gives a remainder of 1 when divided by 2.

### Finding the possible values of \( N \):
We need to find a number that satisfies both conditions.

1. From \( N \equiv 3 \ (\text{mod} \ 10) \), we know \( N \) can be expressed as:
   \[
   N = 10k + 3
   \]
   for some integer \( k \).

2. Now, substitute this into the second condition, \( N \equiv 1 \ (\text{mod} \ 2) \):
   \[
   10k + 3 \equiv 1 \ (\text{mod} \ 2)
   \]
   Simplifying:
   \[
   (10k \mod 2) + (3 \mod 2) \equiv 1 \ (\text{mod} \ 2)
   \]
   Since \( 10k \mod 2 = 0 \) (because 10 is even) and \( 3 \mod 2 = 1 \), we get:
   \[
   0 + 1 \equiv 1 \ (\text{mod} \ 2)
   \]
   This is true, so no further conditions are needed.

### Conclusion:
The possible values of \( N \) are all numbers of the form \( N = 10k + 3 \), where \( k \) is an integer. The sequence of possible values is:
\[
N = 3, 13, 23, 33, 43, 53, 63, 73, 83, 93, 103, 113, \dots
\]

These numbers are spaced by 10 units, and there are infinitely many possible solutions. The number of chocolates you can buy depends on how much space is available in the handbag.

---

Would you like more details or have any questions about this solution?

Here are 5 related questions to deepen your understanding:
1. What does it mean for a number to be congruent modulo another number?
2. Can you find another method to solve for \( N \) given two modular conditions?
3. How would the solution change if there were 2 chocolates left over when counted in groups of 5?
4. Could we solve this problem with only the condition \( N \equiv 3 \ (\text{mod} \ 10) \)?
5. What is the smallest \( N \) such that both conditions hold and \( N > 100 \)?

**Tip:** In problems with remainders like this, look for patterns in the multiples and solve them using modular arithmetic.

A chocolate shop is closing down and needs to sell all their chocolates. Your Mum only allows you to buy enough to fit in her handbag. The chocolates were counted in groups of 10 but there were 3 left over. The chocolates were counted in groups of 2 but there was 1 left over. How many chocolates could you buy? How many possible answers can you come up with?

Learn how to solve a fun kids' math problem involving modular arithmetic to find the number of chocolates. Understand key concepts such as congruences and the Chinese Remainder Theorem. Suitable for students in Grades 6-8.

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