This worksheet, titled 'Problems and More,' offers a variety of brain teasers including multiplication challenges, money sums, and logic puzzles.

Let's solve each problem step by step:

---

Problem 1: Getting to 360

Task: Write two different multiplication problems that have a product of 360.

#### Solution:
To find two different multiplication problems with a product of 360, we need to factorize 360 and create pairs of factors.

1. Factorization of 360:
- \( 360 = 2^3 \times 3^2 \times 5 \)

2. Finding Factor Pairs:
- Some possible factor pairs of 360 are:
- \( 360 = 18 \times 20 \)
- \( 360 = 12 \times 30 \)
- \( 360 = 9 \times 40 \)
- \( 360 = 6 \times 60 \)
- \( 360 = 5 \times 72 \)
- \( 360 = 4 \times 90 \)
- \( 360 = 3 \times 120 \)
- \( 360 = 2 \times 180 \)
- \( 360 = 1 \times 360 \)

3. Choosing Two Different Problems:
- Let's choose two distinct pairs:
- \( 18 \times 20 = 360 \)
- \( 12 \times 30 = 360 \)

#### Final Answer:
\[
\boxed{18 \times 20 = 360, \quad 12 \times 30 = 360}
\]

---

Problem 2: More and More Money

Task: If you received 1¢ on March 1st, 2¢ on March 2nd, 3¢ on March 3rd, and so on for the entire month of March (31 days), how much money would you have at the end of the month?

#### Solution:
This is an arithmetic series where:
- The first term (\(a\)) is 1¢.
- The common difference (\(d\)) is 1¢.
- The number of terms (\(n\)) is 31 (since March has 31 days).

The sum of an arithmetic series is given by the formula:
\[
S_n = \frac{n}{2} \times (a + l)
\]
where:
- \(S_n\) is the sum of the series,
- \(n\) is the number of terms,
- \(a\) is the first term,
- \(l\) is the last term.

Here:
- \(a = 1\),
- \(l = 31\) (since the last term on March 31st is 31¢),
- \(n = 31\).

Substitute these values into the formula:
\[
S_{31} = \frac{31}{2} \times (1 + 31) = \frac{31}{2} \times 32 = 31 \times 16 = 496
\]

#### Final Answer:
\[
\boxed{496 \text{ cents}}
\]

---

Problem 3: Going in Circles?

Task: Fill in the circles with the numbers from 2 to 6 so that each side of the triangle adds up to 10.

#### Solution:
We need to place the numbers 2, 3, 4, 5, and 6 in the circles such that the sum of the numbers on each side of the triangle is 10.

1. Visual Representation:
```
A
/ \
B---C
/ \ / \
D---E---F
```
- Each side of the triangle is \(A + B + D\), \(B + C + E\), and \(C + F + D\).
- We need \(A + B + D = 10\), \(B + C + E = 10\), and \(C + F + D = 10\).

2. Assigning Numbers:
- Start by placing the largest numbers in the corners (A, C, F) because they will be part of multiple sums.
- Try different combinations to satisfy all three conditions.

After testing different arrangements, one valid solution is:
- \(A = 6\), \(B = 3\), \(C = 5\), \(D = 1\), \(E = 2\), \(F = 4\).

Verify:
- Side 1: \(A + B + D = 6 + 3 + 1 = 10\)
- Side 2: \(B + C + E = 3 + 5 + 2 = 10\)
- Side 3: \(C + F + D = 5 + 4 + 1 = 10\)

#### Final Answer:
\[
\boxed{6, 3, 5, 1, 2, 4}
\]

---

Problem 4: Coin Capers

Task: How many ways can you make 50¢ using at least one penny, one nickel, one dime, and one quarter?

#### Solution:
We need to make 50¢ using at least one of each coin type: penny (1¢), nickel (5¢), dime (10¢), and quarter (25¢).

1. Minimum Coins Required:
- Use one of each coin:
- 1 quarter = 25¢
- 1 dime = 10¢
- 1 nickel = 5¢
- 1 penny = 1¢
- Total = \(25 + 10 + 5 + 1 = 41¢\).

2. Remaining Amount:
- We need \(50 - 41 = 9¢\) more.

3. Ways to Make 9¢ Using Pennies and Nickels:
- Since we already have one nickel, we can only use pennies to make up the remaining 9¢.
- Possible combinations:
- 9 pennies (9 × 1¢ = 9¢).

Thus, there is only one way to make 50¢ using at least one of each coin.

#### Final Answer:
\[
\boxed{1}
\]

---

Problem 5: What's Your Sign?

Task: Fill in the missing + and – signs to make the equation true:
\[
5 \circ 4 \circ 9 \circ 3 \circ 2 \circ 1 = 4
\]

#### Solution:
We need to insert + and – signs between the numbers to make the equation equal to 4.

1. Trial and Error:
- Start by testing different combinations of + and – signs.

One valid combination is:
\[
5 - 4 + 9 - 3 - 2 - 1 = 4
\]

Verify:
\[
5 - 4 = 1, \quad 1 + 9 = 10, \quad 10 - 3 = 7, \quad 7 - 2 = 5, \quad 5 - 1 = 4
\]

#### Final Answer:
\[
\boxed{5 - 4 + 9 - 3 - 2 - 1 = 4}
\]

---

Final Answers:

1. \( \boxed{18 \times 20 = 360, \quad 12 \times 30 = 360} \)
2. \( \boxed{496 \text{ cents}} \)
3. \( \boxed{6, 3, 5, 1, 2, 4} \)
4. \( \boxed{1} \)
5. \( \boxed{5 - 4 + 9 - 3 - 2 - 1 = 4} \)