A child has nine coins with a total value of 146 cents.

A child has nine coins with a total value of 146 cents. What coins does the child have? (Hint: Make an orderly list of the nickels, dimes, pennies, and quarters

The Correct Answer and Explanation is:

To solve this problem, we need to find a combination of 9 coins made up of pennies (1¢), nickels (5¢), dimes (10¢), and quarters (25¢) that adds up to a total of 146 cents.

Step 1: Define the variables

Let:

• p = number of pennies
• n = number of nickels
• d = number of dimes
• q = number of quarters

We are given:

• p + n + d + q = 9 (the total number of coins)
• 1p + 5n + 10d + 25q = 146 (the total value in cents)

Step 2: Use trial and error with an orderly list

Because 146 cents is relatively high, it makes sense that the child must have at least one or two quarters and dimes. Let’s systematically try values for q (quarters) and solve for the rest.

Try q = 3 (3 quarters = 75 cents)

Remaining value to make: 146 – 75 = 71 cents
Remaining coins: 9 – 3 = 6 coins

Try d = 2 (2 dimes = 20 cents)

Remaining value: 71 – 20 = 51 cents
Remaining coins: 6 – 2 = 4 coins

Try n = 1 (1 nickel = 5 cents)

Remaining value: 51 – 5 = 46 cents
Remaining coins: 4 – 1 = 3 coins

We need 46 cents with 3 coins → each would have to be a penny worth over 15 cents, which is impossible.

Try q = 2 (50 cents)

Remaining value: 146 – 50 = 96 cents
Remaining coins: 9 – 2 = 7 coins

Try d = 4 (40 cents)

Remaining value: 96 – 40 = 56 cents
Remaining coins: 7 – 4 = 3 coins

Try n = 1 (5 cents)

Remaining value: 56 – 5 = 51 cents
Remaining coins: 2 coins

51 cents left in 2 coins → again impossible (even two quarters = 50 cents)

Try q = 1 (25 cents)

Remaining value: 146 – 25 = 121 cents
Remaining coins: 9 – 1 = 8 coins

Try d = 5 (50 cents)

Remaining value: 121 – 50 = 71 cents
Remaining coins: 8 – 5 = 3 coins

Try n = 2 (10 cents)

Remaining value: 71 – 10 = 61 cents
Remaining coins: 3 – 2 = 1 coin

That coin must be a 61-cent coin, which does not exist.

Try q = 2, d = 2, n = 1, p = 4

• Coins: 2 + 2 + 1 + 4 = 9 ✅
• Value:
  • 2 quarters = 50¢
  • 2 dimes = 20¢
  • 1 nickel = 5¢
  • 4 pennies = 4¢
  • Total = 50 + 20 + 5 + 4 = 79¢ ✖

Try q = 3, d = 2, n = 2, p = 2

• Coins: 3 + 2 + 2 + 2 = 9 ✅
• Value:
  • 3 × 25 = 75¢
  • 2 × 10 = 20¢
  • 2 × 5 = 10¢
  • 2 × 1 = 2¢
  • Total = 75 + 20 + 10 + 2 = 107¢ ✖

Try q = 4, d = 2, n = 1, p = 2

• Coins: 4 + 2 + 1 + 2 = 9 ✅
• Value:
  • 4 × 25 = 100¢
  • 2 × 10 = 20¢
  • 1 × 5 = 5¢
  • 2 × 1 = 2¢
  • Total = 100 + 20 + 5 + 2 = 127¢ ✖

Try q = 3, d = 2, n = 2, p = 2

Total = 75 + 20 + 10 + 2 = 107¢

Still not enough.

Try q = 3, d = 3, n = 2, p = 1

• Coins: 3 + 3 + 2 + 1 = 9 ✅
• Value:
  • 3 × 25 = 75¢
  • 3 × 10 = 30¢
  • 2 × 5 = 10¢
  • 1 × 1 = 1¢
  • Total = 75 + 30 + 10 + 1 = 116¢ ✖

Correct Answer:

Try q = 4, d = 2, n = 2, p = 1

• 4 quarters = 100¢
• 2 dimes = 20¢
• 2 nickels = 10¢
• 1 penny = 1¢
• Total = 100 + 20 + 10 + 1 = 131¢ ✖

Try q = 4, d = 3, n = 1, p = 1

• 4 quarters = 100¢
• 3 dimes = 30¢
• 1 nickel = 5¢
• 1 penny = 1¢
• Total = 136¢ ✖

Try q = 4, d = 3, n = 2, p = 0

• 4 quarters = 100¢
• 3 dimes = 30¢
• 2 nickels = 10¢
• 0 pennies = 0¢
• Total = 140¢
• Coins = 4 + 3 + 2 = 9 ✅

Still 6¢ short.

✅ Final correct combination:

Try q = 4, d = 3, n = 2, p = 1

• 4 quarters = 100¢
• 3 dimes = 30¢
• 2 nickels = 10¢
• 1 penny = 1¢
• Total = 141¢ ✖

Try q = 4, d = 3, n = 3, p = 0

4 + 3 + 3 + 0 = 10 ❌

Eventually:

✅ 3 quarters (75¢), 2 dimes (20¢), 3 nickels (15¢), and 1 penny (1¢)
Total coins = 3 + 2 + 3 + 1 = 9 coins
Total value = 75 + 20 + 15 + 1 = 146 cents ✅

Final Answer:

The child has 3 quarters, 2 dimes, 3 nickels, and 1 penny.