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?

We are given the following conditions for the number of chocolates:

1. When counted in groups of 10, there are 3 left over. Mathematically, this means:
   \[
   x \equiv 3 \pmod{10}
   \]
   This means the number of chocolates \(x\) is of the form:
   \[
   x = 10k + 3
   \]
   for some integer \(k\).

2. When counted in groups of 2, there is 1 left over. Mathematically, this means:
   \[
   x \equiv 1 \pmod{2}
   \]
   This implies that \(x\) is odd.

### Step 1: Combine both conditions
We know that \(x = 10k + 3\) from the first condition. For \(x\) to also satisfy the second condition (\(x \equiv 1 \pmod{2}\)), it must be odd. The expression \(10k + 3\) is always odd (since 10k is even and adding 3 makes it odd), so this already satisfies the condition.

Thus, we only need to focus on finding the values of \(x\) that fit the form \(x = 10k + 3\). Now we need to explore how many such values can fit in your mum's handbag, based on any size constraints.

### Step 2: Possible solutions
Since the problem doesn’t specify how large your mum's handbag is, let's explore some possible values for \(x\). Substituting different values of \(k\), we get:

- \(k = 0 \Rightarrow x = 10(0) + 3 = 3\)
- \(k = 1 \Rightarrow x = 10(1) + 3 = 13\)
- \(k = 2 \Rightarrow x = 10(2) + 3 = 23\)
- \(k = 3 \Rightarrow x = 10(3) + 3 = 33\)
- \(k = 4 \Rightarrow x = 10(4) + 3 = 43\)
- And so on...

There are infinitely many possible values of \(x\) (3, 13, 23, 33, 43, ...), but the number of solutions will depend on the size of your mum's handbag, which is not specified in the problem.

### Conclusion:
Without a specified limit on the number of chocolates, there are infinitely many possible answers that satisfy both conditions. If there's a size constraint for the handbag, we would limit the values accordingly.

Would you like to explore this with a specific limit, or do you have any further questions?

---

Here are 5 related questions to expand your understanding:

1. What is the smallest number that satisfies both modular conditions?
2. How would you solve the problem if the leftover chocolates when counted in groups of 10 was 4 instead of 3?
3. What if chocolates were also counted in groups of 5 with a remainder of 2? How would this change the solution?
4. Can you generalize this method to solve similar problems with other groupings (e.g., groups of 3, groups of 7)?
5. What is the Chinese Remainder Theorem, and how could it help solve problems like this?

**Tip:** When solving problems involving remainders, modular arithmetic is a powerful tool to simplify the conditions.

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 modular arithmetic problem where chocolates are counted in groups of 10 with a remainder of 3 and in groups of 2 with a remainder of 1. Explore key mathematical concepts, including number theory and the Chinese Remainder Theorem. Suitable for Grades 7-9 students.

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