55 Cents, Seven CoinsDate: 05/15/2001 at 17:08:42 From: Erik Subject: Money How can you get 55 cents with seven coins (quarters, dimes, nickels or pennies only) eight ways? Date: 05/16/2001 at 15:40:36 From: Doctor TWE Subject: Re: Money Hi Erik - thanks for writing to Dr. Math. I only count three ways of doing it - and more importantly, I can prove there aren't any other solutions. Let's start by thinking about how we can simplify this problem. Pennies aren't worth very much and 'use' a lot of coins, so let's see if we can't eliminate some combinations. Since the amount we are trying to get is a multiple of 5 cents, and since all other coins (nickels, dimes, and quarters) are also multiples of 5 cents, any solution involving pennies would have to use a multiple of 5. Otherwise we'd have a few cents 'left over'. We are only allowed seven coins total, and the only multiples of 5 that are less than 7 are 5 and 0, so our solutions have to have either five pennies or zero pennies. Let's consider solutions with five pennies. If five of the coins are pennies, then the other 7-5 = 2 coins must be worth 55-5 = 50 cents. The only way to get 50 cents with exactly two coins is with two quarters. So that's one solution: 2 quarters and 5 pennies Furthermore, that's the ONLY solution that uses pennies. (We know all solutions must have a multiple of five pennies and can't have more than seven pennies; and we know that that is the only solution with exactly five pennies. Thus all other solutions must have no pennies.) Now we have to find solutions that involve only nickels, dimes, and quarters. Can we narrow it down more? Let's look at the other extreme - quarters. The solutions obviously can't have more than two quarters (otherwise the total would be more than 55 cents). Let's consider solutions with exactly two quarters. If two of the coins are quarters, then the other 7-2 = 5 coins must be worth 55-50 = 5 cents. The only way to get 5 cents with exactly five coins is with five pennies. But that's the solution we already came up with. Using only nickels and dimes, there's no way to get 5 cents using exactly five coins. So we know that any further solutions must have either one quarter or no quarters. Let's next consider solutions with exactly one quarter. If one of the coins is a quarter, then the other 7-1 = 6 coins must be worth 55-25 = 30 cents. (And we know that there are no pennies involved.) Can you come up with ways of getting 30 cents with exactly six coins that are nickels or dimes only? Furthermore, can you *prove* that you've found all solutions? Finally, we consider the case of no quarters (and no pennies). Can you come up with ways of getting 55 cents with exactly seven coins that are nickels or dimes only? Can you prove that you've found all solutions? I'll let you finish the work on those last two cases. Once you've found them, you're done. I hope this helps. If you have any more questions, write back. - Doctor TWE, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/