Name: Rachelle
Level: Middle 7th grade
asking: Student


What is the least positive number of coins that is impossible to give out change for a dollar? It is higher than 50 i know and you can us dimes, nickels, pennies, quarters, 50 cent pieaces and silver dollars.

please help as soon as possible

Hi Rachelle,

Suppose that you have change for a dollar using some number of coins (call it N), but that there is no conceivable way to give correct change for a dollar with one more coin.

Then you have no 50 cents piece, because exchanging it for two quarters would give change for a dollar with one more coin.

Similarly, you have no dime, because it is two nickles.

You do not have three quarters, because it is a 50 cents piece, two dimes and a quarter,

You do not have both a quarter and nickel because it is three dimes.

You cannot have seven nickels, because it is three dimes and five pennies.

So your options are:

  1. Two quarters, 50 pennies: N = 52 (N+1 = 53)
  2. One quarter, 75 pennies: N = 76 (N+1 = 77)
  3. 100 pennies, N = 100 (N+1 = 101)
  4. 6 nickels, 70 pennies: N = 76 (N+1 = 77)
  5. 5 nickels, 75 pennies: N = 80 (N+1 = 81)
  6. 4 nickels, 80 pennies: N = 84 (N+1 = 85)
  7. 3 nickels, 85 pennies: N = 88 (N+1 = 89)
  8. 2 nickels, 90 pennies: N = 92 (N+1 = 93)
  9. 1 nickel, 95 pennies: N = 96 (N+1 = 97)

However a) is not the solution because you can change a dollar with three dimes, five quarters and 45 pennies: 53 pieces

b) and d) are the correct answer: Suppose that you have the correct change for a dollar with 77 pieces. Then the number of pennies HAS to be a multiple of 5. (Can you figure out why?) If it were 75, then the two other coins would have to add up to 25 cents, which is impossible. If it were 70 or less, then even if the other coins were all nickels, the total amount would add up to more than one dollar.

Go to Math Central