25 Coins in a DollarDate: 08/16/2001 at 10:58:46 From: Sheila Williams Subject: How to get a dollar using 25 coins. My son's 6th grade teacher gave him a puzzle to try in the classroom and he was very discouraged when several of the students got the answer right away. His teacher told him to try it at home. The question was, "How can you get a dollar using 25 coins?" She told him that there were many different combinations but to find only five of them. Is it just trial and error or is there a formula that can be worked out like (ax1) + (bx5) + (cx10) + (dx25) = one dollar? Thank you. Sheila Williams Date: 08/16/2001 at 15:53:23 From: Doctor Rob Subject: Re: How to get a dollar using 25 coins. Thanks for writing to Ask Dr. Math, Sheila. If you have Q quarters, D dimes, N nickels, and P pennies, then you want nonnegative integer values of these variables that satisfy 25*Q + 10*D + 5*N + P = 100 (total value of coins), Q + D + N + P = 25 (total number of coins). Subtracting one from the other, you have 24*Q + 9*D + 4*N = 75. Since 3 is a divisor of 24, 9, and 75, it must be a divisor of N. Put N = 3*X. Then, substituting and dividing by 3, you get 8*Q + 3*D + 4*X = 25 = 16 + 9. Now since 8 and 4 and 16 are divisible by 4, 3*D - 9 = 3*(D-3) must also be divisible by 4. That means that D - 3 must be divisible by 4, so D = 4*Y + 3. Substituting, combining like terms, and dividing by 4, you get 2*Q + 3*Y + X = 4. Now you can explicitly write down all the nonnegative integer solutions to this equation, and then calculate the values of D = 4*Y + 3, N = 3*X, and P = 25 - Q - D - N: Q Y X D N P 0 0 4 3 12 10 1 0 2 3 6 15 2 0 0 3 0 20 0 1 1 7 3 15 You can always trade 1 quarter and 5 pennies for 6 nickels and vice versa, also 4 dimes and 5 pennies for 9 nickels. That gives you four solutions. To get more than four, you need half-dollars. Then if you have one half-dollar, the equations become 25*Q + 10*D + 5*N + P = 50, Q + D + N + P = 24, 24*Q + 9*D + 4*N = 26 = 8 + 18. Then 9*(D-2), and so D - 2, must be divisible by 4, so D must be of the form 4*Y + 2, so that 6*Q + 9*Y + N = 2. The only solution to this is Q = Y = 0, N = 2, so D = 2 and P = 20. This means one half-dollar, no quarters, two dimes, two nickels, and 20 pennies. You can always trade a half-dollar and 5 pennies for 5 dimes and a nickel. That is the fifth solution. There are only five. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 08/16/2001 at 18:20:19 From: Doctor Greenie Subject: Re: How to get a dollar using 25 coins. Hi, Sheila -- Dr. Rob has written a detailed response to your question, but while it's fine for a talented high school student, it may not help a 6th-grader work through the problem. Here is how I would do this problem with a 6th-grader: There are two keys to my approach to this type of problem: (*) Nickels, dimes, quarters, and half dollars together can only make total numbers of cents that are multiples of 5. If the total of 25 coins is to be exactly one dollar (100 cents), then the number of pennies must be a multiple of 5. (**) With "n" coins, all of which are either nickels or dimes, you can make any total number of cents that is between 5n (all nickels) and 10n (all dimes). For example, suppose you have 4 coins that are all either nickels or dimes. The smallest amount you can make happens if all four coins are nickels; the total is 20 cents. Now trade one of those nickels for a dime. You lose the 5 cents from the nickel and gain 10 cents from the dime, for a net gain of 5 cents, so now your total is 25 cents. Trade a second nickel for a second dime, and you increase your total amount by another 5 cents, to 30 cents. Trade a third nickel for a third dime and you have a total of 35 cents; and trade the last nickel for a dime and you have a total of 40 cents. With (*) and (**) above in mind, the approach to solving a problem like this is the following: (1) Choose a number of half dollars and/or quarters that keeps the total amount in half dollars and quarters under 100 cents. (2) For each combination of numbers of half dollars and quarters, determine what numbers of pennies you can have (the number of pennies must be a multiple of 5; and the total number of coins among half dollars, quarters, and pennies must be less than 25). (3) For each combination of half dollars, quarters, and pennies, determine the number of coins remaining to make 25 coins and the number of cents remaining to make 100 cents. According to (**) above, there is a solution to the problem using this combination of half dollars, quarters, and pennies if (and only if) the number of cents remaining to be made is between 5 and 10 times the number of coins remaining. The process is much easier to use than it is to explain. Here is the analysis for this particular problem that shows the five possible solutions. # coins # cents 50c 25c 1c remaining remaining solution? --------------- -------------------- ------------------------ 1 1 20 3 5 no 1 1 15 8 10 no 1 1 10 13 15 no 1 1 5 18 20 no 1 1 0 23 25 no 1 0 20 4 30 yes (2 dimes, 2 nickels) 1 0 15 9 35 no 1 0 10 14 40 no 1 0 5 19 45 no 1 0 0 24 50 no 0 3 20 2 5 no 0 3 15 7 10 no 0 3 10 12 15 no 0 3 5 17 20 no 0 3 0 22 25 no 0 2 20 3 30 yes (3 dimes) 0 2 15 8 35 no 0 2 10 13 40 no 0 2 5 18 45 no 0 2 0 23 50 no 0 1 20 4 55 no 0 1 15 9 60 yes (3 dimes, 6 nickels) 0 1 10 14 65 no 0 1 5 19 70 no 0 1 0 24 75 no 0 0 20 5 80 no 0 0 15 10 85 yes (7 dimes, 3 nickels) 0 0 10 15 90 yes (3 dimes, 12 nickels) 0 0 5 20 95 no 0 0 0 25 100 no - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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