Simply breakdown for children : 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?

Alright, let's break this down step by step with some fun!

Imagine you go to the chocolate shop, and they're about to close. You can only buy enough chocolates to fit in your Mum's handbag. But how many chocolates could that be? Here's what we know:

1. When the chocolates were counted in groups of 10, there were **3 left over**. This means that if you count the total chocolates by 10, you always get 3 left. So, the number of chocolates could look like this:
   - 3, 13, 23, 33, 43, 53, and so on (all numbers that end with 3 when divided by 10).

2. When the chocolates were counted in groups of 2, there was **1 left over**. This means that if you count the total chocolates by 2, you always get 1 left. So the number of chocolates could look like this:
   - 1, 3, 5, 7, 9, 11, and so on (all odd numbers).

Now, we need to figure out which numbers fit both clues. We’re looking for numbers that are:
- Left over 3 when counted by 10 (like 3, 13, 23, 33, 43…)
- Left over 1 when counted by 2 (odd numbers like 1, 3, 5, 7…)

From this, we can see that **3** fits both clues! It's a number that:
- Ends with 3 when divided by 10.
- Is odd, so it leaves 1 when divided by 2.

So the possible answer is 3 chocolates.

But is that the only answer? Let’s see if there are other numbers that also work. We can try numbers like 13, 23, 33, 43, 53... to see if they fit both clues.

**Possible answers:**
- 3, 23, 43, 63, 83, 103, and so on...

You could come up with **many answers**, as long as the numbers fit both rules.

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Do you want more details or have any questions? Here are 5 questions to expand on this idea:

1. Why do we get 3 when we divide by 10 with some leftover?
2. What happens if we count the chocolates in groups of 5?
3. How can we check for other numbers that fit both clues?
4. Why are some numbers left over when divided?
5. Can you find the next number after 103 that fits the clues?

**Tip:** When solving these kinds of problems, try listing numbers that fit one clue, then see if they fit the second clue.

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?

Discover how to solve a problem using modular arithmetic where chocolates are counted in groups of 10 with a remainder of 3, and in groups of 2 with a remainder of 1. This step-by-step solution introduces number theory concepts and is suitable for students in Grades 5-7.

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