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21 Nickels, Dimes, and Pennies in a Dollar

From: Mary C. DuChateau
Subject: Coin question
Date: Tue, 8 Nov 94 20:15:23 EST

Hello Dr. math-

I am showing my eleven year-old electronic mail
on the Internet, and he also has a math question for school
homework.  We are going to "kill two birds with one stone."

His question is: How can you have 21 coins of nickels, dimes, 
and pennies, that equal one dollar using no quarters?  

Please help us out.  Thank you so much -

Marie and John DuChateau

Date: Tue, 8 Nov 1994 20:55:15 -0500
From: Doctor Melissa
Subject: Re: Coin question

Hello!  I'm not sure how much math John has had, but I'll try to 
show this using algebra.  If he needs more of an explanation, 
please write back!

We're trying to find what number of coins of pennies nickels and 
dimes add up to $1.00.

We can rewrite this equation as:

0.10D + 0.05N + 0.01P = 1.00

Multiply everything by 100 to make the numbers easier to read on 
a computer:

10D + 5N + P = 100

where D is the number of dimes, N is the number of nickels, and 
P is the number of pennies.

But that isn't enough information to solve the problem - we were 
also told that there are 21 coins.  So we have another equation:

D + N + P = 21

We can then rewrite it as:

P = 21 - D - N

and substitute this into the original equation to get:

10D + 5N + (21 - D - N) = 100

Which is equal to:

9D + 4N + 21 = 100


9D + 4N = 79

Since there are two variables here, we know there are many 
different answers. To find an answer with only integers (no 
fractions) in it, you can either graph it or just try values for 
D (Hint: there is a value less than 5 that gives a counting 
number answer for N) and solve for N.

Good luck!

Dr. Melissa

Date: 6/25/96 at 13:27:30
From: Doctor Gary
Subject: Re: Coin question
Here's another answer.

Since 20 nickels are worth a dollar, at least one of 
the coins must be worth less than five cents, so there 
has to be at least one penny.

If there's at least one penny, there have to be at 
least 5 pennies, because the value of all the coins 
(like the value of each nickel and dime) is a multiple 
of five cents.

So, the possible numbers of pennies are five, ten, 
fifteen, and twenty.  We can rule out fifteen and 
twenty, because all the other coins (even if they were 
all dimes) wouldn't be worth enough to make up the 
difference between the value of the pennies and one 

So we've only to consider two possible scenarios:

   (1)  5 pennies, 16 nickels and dimes

       16 nickels are worth 80 cents.  "Trade" three 
nickels for dimes and you've got 13 nickels (worth 65 
cents) and 3 dimes (worth 30 cents), to go along with 
your five pennies.

   (2) 10 pennies, 11 nickels and dimes

       11 nickels are worth 55 cents.  "Trade" 7 nickels 
for dimes and you've got 4 nickels (worth 20 cents) and 
7 dimes (worth 70 cents) to go along with your ten 

So the two solutions to the problem are:
5 pennies, 13 nickels, 3 dimes,  or
10 pennies, 4 nickels, 7 dimes.

Math is just common sense with a pencil.  Sometimes, 
common sense is even faster without the pencil. 

-Doctor Gary,  The Math Forum
 Check out our web site!   

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