Ways To Pay $2.90 Using Coins A Mathematical Analysis

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This article dives into an intriguing mathematical problem: determining the number of ways to pay $2.90 using only 10-cent, 20-cent, and 50-cent coins, with the constraint of using at least four of each coin. This problem blends basic arithmetic with combinatorial thinking, offering a great way to explore problem-solving strategies. Let's embark on this mathematical journey and uncover the solutions!

Problem Statement

The core of our challenge lies in figuring out how many different combinations of 10-cent, 20-cent, and 50-cent coins can add up to $2.90, but there's a twist. We need to use at least four of each coin denomination. This condition adds an interesting layer of complexity to the problem, making it more than just a straightforward sum calculation. It requires us to think systematically about how different coin combinations interact and how the 'at least four' rule affects our options. This problem is a perfect example of how real-world scenarios can translate into engaging mathematical puzzles.

Breaking Down the Problem

To effectively tackle this problem, we'll first convert the dollar amount into cents to simplify our calculations. $2.90 becomes 290 cents. We know we must use at least four of each coin type, so let's calculate the minimum amount contributed by these coins: Four 10-cent coins equal 40 cents, four 20-cent coins equal 80 cents, and four 50-cent coins equal 200 cents. Adding these together, the minimum total is 40 + 80 + 200 = 320 cents. However, this is already more than our target of 290 cents! This tells us we can't use the minimum of all three coin types simultaneously. This is a crucial insight that will guide our approach.

Now, the puzzle becomes: how can we adjust the number of each coin type to reach exactly 290 cents, keeping in mind the constraint of using at least four of some coins? This is where we'll start exploring different scenarios, adjusting the quantities of each coin type and checking if they add up correctly. The key is to be organized and methodical in our approach to avoid missing any possible combinations.

Solving the Problem: Methodical Approach

Given the minimum coin requirement leads to an excess of 320 cents, exceeding our target of 290 cents, we must reduce the number of coins used. Since we have to use at least four of each coin type initially, the excess tells us we need to reduce the number of 50-cent coins. Let's start by considering scenarios with fewer than four 50-cent coins.

Scenario 1: Reducing 50-Cent Coins

Let's explore what happens if we use fewer than four 50-cent coins. Since we must use at least four of each coin if we use them, the only way to reduce the amount contributed by 50-cent coins is to use zero of them. If we use zero 50-cent coins, we need to make 290 cents using only 10-cent and 20-cent coins, with a minimum of four of each. Four 10-cent coins are 40 cents, and four 20-cent coins are 80 cents, totaling 120 cents. We still need to make up 290 - 120 = 170 cents. Now, we need to find combinations of additional 10-cent and 20-cent coins that sum to 170 cents.

To find these combinations, let's denote the number of additional 10-cent coins as 'x' and the number of additional 20-cent coins as 'y'. We need to solve the equation 10x + 20y = 170. Dividing the equation by 10, we get x + 2y = 17. Now, we can explore possible integer solutions for x and y. Remember, x and y represent additional coins, so we just need non-negative integers.

  • If y = 0, then x = 17. This means 17 additional 10-cent coins. So, we have 17 + 4 = 21 10-cent coins and 4 20-cent coins. This is one valid solution.
  • If y = 1, then x = 15. This means 15 additional 10-cent coins and 1 additional 20-cent coin. So, we have 15 + 4 = 19 10-cent coins and 1 + 4 = 5 20-cent coins. This is another valid solution.
  • If y = 2, then x = 13. This means 13 additional 10-cent coins and 2 additional 20-cent coins. So, we have 13 + 4 = 17 10-cent coins and 2 + 4 = 6 20-cent coins. This is a valid solution.
  • If y = 3, then x = 11. This means 11 additional 10-cent coins and 3 additional 20-cent coins. So, we have 11 + 4 = 15 10-cent coins and 3 + 4 = 7 20-cent coins. This is a valid solution.
  • If y = 4, then x = 9. This means 9 additional 10-cent coins and 4 additional 20-cent coins. So, we have 9 + 4 = 13 10-cent coins and 4 + 4 = 8 20-cent coins. This is a valid solution.
  • If y = 5, then x = 7. This means 7 additional 10-cent coins and 5 additional 20-cent coins. So, we have 7 + 4 = 11 10-cent coins and 5 + 4 = 9 20-cent coins. This is a valid solution.
  • If y = 6, then x = 5. This means 5 additional 10-cent coins and 6 additional 20-cent coins. So, we have 5 + 4 = 9 10-cent coins and 6 + 4 = 10 20-cent coins. This is a valid solution.
  • If y = 7, then x = 3. This means 3 additional 10-cent coins and 7 additional 20-cent coins. So, we have 3 + 4 = 7 10-cent coins and 7 + 4 = 11 20-cent coins. This is a valid solution.
  • If y = 8, then x = 1. This means 1 additional 10-cent coin and 8 additional 20-cent coins. So, we have 1 + 4 = 5 10-cent coins and 8 + 4 = 12 20-cent coins. This is a valid solution.

So, in this scenario, we've found 9 different ways to make $2.90 using only 10-cent and 20-cent coins, ensuring we use at least four of each if they are used.

Scenario 2: Using One 50-Cent Coin

Now, let's consider the case where we use one 50-cent coin. This contributes 50 cents to the total, leaving us with 290 - 50 = 240 cents to make up. We still need to use at least four 10-cent and four 20-cent coins, which account for 40 cents and 80 cents, respectively, totaling 120 cents. Subtracting this from the remaining amount, we have 240 - 120 = 120 cents left to make up using additional 10-cent and 20-cent coins.

Let's again use 'x' for additional 10-cent coins and 'y' for additional 20-cent coins. Our equation now is 10x + 20y = 120. Dividing by 10, we get x + 2y = 12. We need to find non-negative integer solutions for x and y:

  • If y = 0, then x = 12. This gives us 12 + 4 = 16 10-cent coins and 4 20-cent coins. This is a valid solution.
  • If y = 1, then x = 10. This gives us 10 + 4 = 14 10-cent coins and 1 + 4 = 5 20-cent coins. This is a valid solution.
  • If y = 2, then x = 8. This gives us 8 + 4 = 12 10-cent coins and 2 + 4 = 6 20-cent coins. This is a valid solution.
  • If y = 3, then x = 6. This gives us 6 + 4 = 10 10-cent coins and 3 + 4 = 7 20-cent coins. This is a valid solution.
  • If y = 4, then x = 4. This gives us 4 + 4 = 8 10-cent coins and 4 + 4 = 8 20-cent coins. This is a valid solution.
  • If y = 5, then x = 2. This gives us 2 + 4 = 6 10-cent coins and 5 + 4 = 9 20-cent coins. This is a valid solution.
  • If y = 6, then x = 0. This gives us 4 10-cent coins and 6 + 4 = 10 20-cent coins. This is a valid solution.

In this scenario, we've found 7 different ways to make $2.90 using one 50-cent coin, along with 10-cent and 20-cent coins, maintaining the minimum coin requirement.

Scenario 3: Using Two 50-Cent Coins

Let's explore the scenario where we use two 50-cent coins. This contributes 100 cents, leaving us with 290 - 100 = 190 cents to make up. We still need at least four 10-cent and four 20-cent coins, totaling 120 cents. Subtracting this, we have 190 - 120 = 70 cents remaining to be made up with additional 10-cent and 20-cent coins.

Again, let's use 'x' for additional 10-cent coins and 'y' for additional 20-cent coins. Our equation is now 10x + 20y = 70. Dividing by 10, we get x + 2y = 7. Let's find the non-negative integer solutions:

  • If y = 0, then x = 7. This gives us 7 + 4 = 11 10-cent coins and 4 20-cent coins. This is a valid solution.
  • If y = 1, then x = 5. This gives us 5 + 4 = 9 10-cent coins and 1 + 4 = 5 20-cent coins. This is a valid solution.
  • If y = 2, then x = 3. This gives us 3 + 4 = 7 10-cent coins and 2 + 4 = 6 20-cent coins. This is a valid solution.
  • If y = 3, then x = 1. This gives us 1 + 4 = 5 10-cent coins and 3 + 4 = 7 20-cent coins. This is a valid solution.

In this scenario, we've found 4 different ways to make $2.90 using two 50-cent coins, along with 10-cent and 20-cent coins, adhering to the minimum coin requirement.

Scenario 4: Using Three 50-Cent Coins

Finally, let's consider the case where we use three 50-cent coins. This contributes 150 cents, leaving us with 290 - 150 = 140 cents to make up. We still need at least four 10-cent and four 20-cent coins, totaling 120 cents. Subtracting this, we have 140 - 120 = 20 cents remaining. To make up this remaining 20 cents, we can use one additional 20-cent coin.

  • So, we have 4 10-cent coins and 4 + 1 = 5 20-cent coins. This is only 1 valid solution.

Total Number of Ways

Now, let's add up the number of ways we found in each scenario:

  • Scenario 1 (Zero 50-cent coins): 9 ways
  • Scenario 2 (One 50-cent coin): 7 ways
  • Scenario 3 (Two 50-cent coins): 4 ways
  • Scenario 4 (Three 50-cent coins): 1 way

Therefore, the total number of ways to pay $2.90 using 10-cent, 20-cent, and 50-cent coins, with at least four of each if used, is 9 + 7 + 4 + 1 = 21 ways.

Conclusion

In conclusion, there are 21 different ways to pay $2.90 using 10-cent, 20-cent, and 50-cent coins, given the constraint of using at least four of each denomination. This problem demonstrates how a systematic and methodical approach can break down a seemingly complex challenge into manageable steps. By considering different scenarios and using basic algebraic equations, we were able to find all possible solutions. This exercise not only provides a solution to a specific mathematical puzzle but also illustrates the broader application of problem-solving strategies in various contexts.

Ways to Pay: In this mathematical exploration, we focused on determining the various ways to pay a specific amount using a limited set of coins.

$2.90 with Coins: The central problem revolves around finding the number of combinations to make $2.90 with coins of 10-cent, 20-cent, and 50-cent denominations.

Mathematical Problem: This article addresses a specific mathematical problem that involves combinatorics and basic arithmetic.

Combinatorial Thinking: Solving this problem requires combinatorial thinking to explore different coin combinations.

Problem-Solving Strategies: Throughout the solution, we employed various problem-solving strategies to systematically break down the challenge.